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

typedef long long ll;

const long long mod=1000000007ll;
void ad(long long &a,long long b){a+=b;a%=mod;}
void mn(long long &a,long long b){a+=mod-b;a%=mod;}
void ml(long long &a,long long b){a*=b;a%=mod;}

int main()
{
  int N,K;
  scanf("%d%d",&N,&K);
  static ll P[200];
  P[0]=1ll;
  for(int i=0;i<=N;i++){
    P[i+1]=P[i]*(K+1)%mod;
  }
  static ll dp[10010][200]={0};
  ll ans=0ll;
  dp[0][N]=1;
  for(int i=N-1;i>=0;i--){
    for(int y=0;y<=N*N;y++){
      if(!dp[y][i+1])continue;
      //printf("%d %d %lld\n",i,y,dp[y][i+1]);
      for(int x=0;x<=K;x++){
	if(x<=i+1&&i+1-x<=y){
	  int p=(x+y)/(i+1);
	  ad(ans,P[i]*p%mod*dp[y][i+1]%mod);
	  ad(dp[y+p][i],dp[y][i+1]);
	}
	else{
	  ad(dp[y][i],dp[y][i+1]);
	}
      }
    }
  }
  printf("%lld\n",(K*(K+1)/2*P[N-1]%mod*N%mod+mod-ans)%mod);
  return 0;
}


### Prompt

In Cpp, your task is to solve 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;

typedef long long ll;

const long long mod=1000000007ll;
void ad(long long &a,long long b){a+=b;a%=mod;}
void mn(long long &a,long long b){a+=mod-b;a%=mod;}
void ml(long long &a,long long b){a*=b;a%=mod;}

int main()
{
  int N,K;
  scanf("%d%d",&N,&K);
  static ll P[200];
  P[0]=1ll;
  for(int i=0;i<=N;i++){
    P[i+1]=P[i]*(K+1)%mod;
  }
  static ll dp[10010][200]={0};
  ll ans=0ll;
  dp[0][N]=1;
  for(int i=N-1;i>=0;i--){
    for(int y=0;y<=N*N;y++){
      if(!dp[y][i+1])continue;
      //printf("%d %d %lld\n",i,y,dp[y][i+1]);
      for(int x=0;x<=K;x++){
	if(x<=i+1&&i+1-x<=y){
	  int p=(x+y)/(i+1);
	  ad(ans,P[i]*p%mod*dp[y][i+1]%mod);
	  ad(dp[y+p][i],dp[y][i+1]);
	}
	else{
	  ad(dp[y][i],dp[y][i+1]);
	}
      }
    }
  }
  printf("%lld\n",(K*(K+1)/2*P[N-1]%mod*N%mod+mod-ans)%mod);
  return 0;
}


```