#include<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int mod=1000000007;

int N,K;

int dp[111][22222];

inline void add(int &a,int b){
    a+=b;
    if(a>=mod)a-=mod;
}

signed main(){
    cin>>N>>K;

    dp[N][0]=1;
    for(int i=N-1;i>=0;i--){
        for(int j=0;j<=K;j++){
            for(int k=0;k<11111;k++){
                if(dp[i+1][k]==0)continue;
                if(j>i+1){
                    add(dp[i][k],dp[i+1][k]);
                }
                else{
                    add(dp[i][k+(j+k)/(i+1)],dp[i+1][k]);
                }
            }
        }
    }

    int sum=0;
    for(int i=0;i<11111;i++)sum=(sum+dp[0][i]*i)%mod;

    int x=N*K*(K+1)/2;
    x%=mod;
    for(int i=0;i<N-1;i++)x=x*(K+1)%mod;

    cout<<(x-sum+mod)%mod<<endl;
    return 0;
}

### Prompt

Generate a CPP solution to the following problem:

Consider the following game:

* The game is played using a row of N squares and many stones.
* First, a_i stones are put in Square i\ (1 \leq i \leq N).
* A player can perform the following operation as many time as desired: "Select an integer i such that Square i contains exactly i stones. Remove all the stones from Square i, and add one stone to each of the i-1 squares from Square 1 to Square i-1."
* The final score of the player is the total number of the stones remaining in the squares.



For a sequence a of length N, let f(a) be the minimum score that can be obtained when the game is played on a.

Find the sum of f(a) over all sequences a of length N where each element is between 0 and K (inclusive). Since it can be extremely large, find the answer modulo 1000000007 (= 10^9+7).

Constraints

* 1 \leq N \leq 100
* 1 \leq K \leq N

Input

Input is given from Standard Input in the following format:


N K


Output

Print the sum of f(a) modulo 1000000007 (= 10^9+7).

Examples

Input

2 2


Output

10


Input

20 17


Output

983853488

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int mod=1000000007;

int N,K;

int dp[111][22222];

inline void add(int &a,int b){
    a+=b;
    if(a>=mod)a-=mod;
}

signed main(){
    cin>>N>>K;

    dp[N][0]=1;
    for(int i=N-1;i>=0;i--){
        for(int j=0;j<=K;j++){
            for(int k=0;k<11111;k++){
                if(dp[i+1][k]==0)continue;
                if(j>i+1){
                    add(dp[i][k],dp[i+1][k]);
                }
                else{
                    add(dp[i][k+(j+k)/(i+1)],dp[i+1][k]);
                }
            }
        }
    }

    int sum=0;
    for(int i=0;i<11111;i++)sum=(sum+dp[0][i]*i)%mod;

    int x=N*K*(K+1)/2;
    x%=mod;
    for(int i=0;i<N-1;i++)x=x*(K+1)%mod;

    cout<<(x-sum+mod)%mod<<endl;
    return 0;
}

```