题解看不懂，不会归并🤪🤪 $O(n^2k\log k)$

首先套路的将$a_{i}$从小到大排序，将求权值改为数 $a_{i+1}-a_{i}$的个数，不难注意到设计$dp$式子为 $f_{i,j,k}$到第$i$个数，分为$j$个集合，首尾确定了$k$个，手玩发现从$i+1$转移到$i$权值变化永远是$(a_{i+1}-a_{i})*(2*j-k)$其实就是在每个集合的的自由端添数，后填的数一定比现在大，为了计算后面的距离所以要加上$(a_{i+1}-a_{i})*(2*j-k)$。 然后就简单了，分类讨论一下是否创建新集合，是否合并集合，是否作为首尾，自己模拟一下即可。码风亲民易看懂

#include<bits/stdc++.h>
#include<unistd.h>
#include<bits/extc++.h>
//#define getchar my_gechar
#define ll long long
#define lll __int128
#define db double
#define ldb long double
#define pb emplace_back
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int> 
#define vec Point
#define fill0(a) memset(a,0,sizeof(a))
#define fill1(a) memset(a,-1,sizeof(a))
#define fillbig(a) memset(a,63,sizeof(a))
// #define max(a,b) ((a)<(b)?(b):(a))
// #define min(a,b) ((a)<(b)?(a):(b))
#define vi vector<int> 
#define vl vector<ll>
#define lowbit(x) (x)&(-x)
//#define sort stable_sort
using namespace std;
using namespace __gnu_pbds;
char buffer[32769];
unsigned li = 0;
inline char my_gechar(){
    if(buffer[li] == '\0') buffer[read(0, buffer, 32768)] = '\0',li = 0;
    return buffer[li++];
}
namespace ly
{
    namespace IO
    {
        #ifndef LOCAL
            constexpr auto maxn=1<<20;
            char in[maxn],out[maxn],*p1=in,*p2=in,*p3=out;

            #define flush() (fwrite(out,1,p3-out,stdout))
            #define putchar(x) (p3==out+maxn&&(flush(),p3=out),*p3++=(x))
            class Flush{public:~Flush(){flush();}}_;
        #endif
        namespace usr
        {
            template<typename type>
                  inline type read(type &x)
            {
                x=0;bool flag(0);char ch=getchar();
                while(!isdigit(ch)) flag^=ch=='-',ch=getchar();
                while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
                return flag?x=-x:x;
            }
            template<typename type>
            inline void write(type x)
            {
                x<0?x=-x,putchar('-'):0;
                static short Stack[50],top(0);
                do Stack[++top]=x%10,x/=10;while(x);
                while(top) putchar(Stack[top--]|48);
            }
            inline int read(double &x){return scanf("%lf",&x);}
            inline int read(long double &x){return scanf("%Lf",&x);}
            inline void dwrite(const double &x,int y=6,bool flag=1){printf("%.*lf",y,x),flag?putchar(' '):putchar(' ');}
            inline void dwrite(const long double &x,int y=6,bool flag=1){printf("%.*Lf",y,x),flag?putchar(' '):putchar(' ');}
            inline char read(char &x){do x=getchar();while(isspace(x));return x;}
            inline char write(const char &x){return putchar(x);}
            inline void read(char *x){static char ch;read(ch);do *(x++)=ch;while(!isspace(ch=getchar())&&~ch);}
            inline void write(const char *x){while(*x)putchar(*(x++));}
            inline void read(string &x){char ch[1000005]={};read(ch),x=ch;}
            inline void write(const string &x){int len=x.length();for(int i=0;i<len;++i)putchar(x[i]);}
            template<typename type,typename...T>
            inline void read(type &x,T&...y){read(x),read(y...);}
            template<typename type,typename...T>
            inline void write(const type &x,const T&...y){write(x),putchar(' '),write(y...),sizeof...(y)^1?0:putchar(' ');}
            inline __int128 read(){static __int128 x;return read(x);}
            template<typename type>
            inline type put(type x,bool flag=1){write(x),flag?putchar(' '):putchar(' ');return x;}
        }
        #ifndef LOCAL
            #undef getchar
            #undef flush
            #undef putchar
        #endif
    }using namespace IO::usr;
}using namespace ly::IO::usr;
struct Point{
    int x,y;
    Point operator +(const Point& b)const{return {x+b.x,y+b.y};}
    Point operator -(const Point& b)const{return {x-b.x,y-b.y};}
    bool operator <(const Point& b)const{
        if(x!=b.x)return x<b.x;
        return y<b.y;
    }
};
const int N=1e5+5;
const int M=1e6+5;
const int inf=1e9+5;
const int mod=1e9+7;
const int bit=50;
//mt19937 sj(chrono::steady_clock::now().time_since_epoch().count());
int B=705;
int n,m;
int a[N];
struct node{
    ll val;
    ll cnt;
    bool operator < (const node &b) const{
        return val>b.val;
    }
};
struct G{
    node a[105];
    int cnt=0;
    inline void init() { cnt=0;}
    inline void new_node(ll x,ll y){
        a[++cnt]={x,y};
    }
    inline void work(){
        sort(a+1,a+cnt+1);
        int tot=0;
        ll sum=0;
        for(int i=1,pos;i<=cnt;i=pos){
            sum=0;
            pos=i;
            while(pos<=cnt&&a[pos].val==a[i].val) (sum+=a[pos++].cnt)%=mod;
            a[++tot]={a[i].val,sum};
            if(tot>=m) break;
        }
        cnt=min(tot,m);
    }
}f[3][2][605];
inline void solve(){
    read(n,m);
    for(int i=1;i<=n;++i) read(a[i]);
    sort(a+1,a+n+1);
    f[0][1][1].new_node(0ll,1ll);
    f[1][1][1].new_node(0ll,2ll);
    int now=0;
    for(int i=n-1;i>=1;--i){
        now^=1;
        for(int j=1;j<=n;++j){
            for(int k=0;k<3;++k){
                if(f[k][now][j].cnt){
                    f[k][now][j].work();
                    for(int t=1;t<=f[k][now][j].cnt;++t){
                        ll val=f[k][now][j].a[t].val+(a[i+1]-a[i])*(2*j-k);
                        ll num=f[k][now][j].a[t].cnt;
                        if(j-k+1>0) f[k][now^1][j+1].new_node(val,num*(j-k+1)%mod);
                        if(2-k>0) f[k+1][now^1][j+1].new_node(val,num*(2-k)%mod);
                        if(j-1>0) f[k][now^1][j-1].new_node(val,num*(j-1)%mod);
                        if(2*j-k>0) f[k][now^1][j].new_node(val,num*(2*j-k)%mod);
                        if(2-k>0) f[k+1][now^1][j].new_node(val,num*(2-k)%mod);
                    }
                    f[k][now][j].init();
                }
            }
        }
    }
    now^=1;
    f[2][now][1].work();
    int i=1;
    for(;i<=min(m,f[2][now][1].cnt);++i){
        write(f[2][now][1].a[i].val,f[2][now][1].a[i].cnt);
        puts("");
    }
    for(;i<=m;++i){
        write(-1,-1);
        puts("");
    }
}

signed main()
{   
    int _=1;
    while(_--) solve();
    return (0-0);
}