#include<bits/stdc++.h>
using namespace std;
using Int = long long;
//INSERT ABOVE HERE
Int dp[111][11111];
Int po[111];
const Int MOD = 1e9+7;
Int dfs(Int n,Int k,Int d){
  if(n==0) return 0;
  Int &res=dp[n][d];
  if(~res) return res;
  res=0;
  for(Int i=0;i<=k;i++){
    if(i<=n) res+=dfs(n-1,k,d+(d+i)/n)+((d+i)%n)*po[n-1];
    else res+=dfs(n-1,k,d)+(d+i)*po[n-1];
    res%=MOD;
  }
  //cout<<n<<" "<<d<<":"<<res<<endl;
  return res;
}
signed main(){
  Int n,k;
  cin>>n>>k;
  
  po[0]=1;
  for(Int i=0;i<n;i++) po[i+1]=po[i]*(k+1)%MOD;
  
  memset(dp,-1,sizeof(dp));
  cout<<dfs(n,k,0)<<endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
using Int = long long;
//INSERT ABOVE HERE
Int dp[111][11111];
Int po[111];
const Int MOD = 1e9+7;
Int dfs(Int n,Int k,Int d){
  if(n==0) return 0;
  Int &res=dp[n][d];
  if(~res) return res;
  res=0;
  for(Int i=0;i<=k;i++){
    if(i<=n) res+=dfs(n-1,k,d+(d+i)/n)+((d+i)%n)*po[n-1];
    else res+=dfs(n-1,k,d)+(d+i)*po[n-1];
    res%=MOD;
  }
  //cout<<n<<" "<<d<<":"<<res<<endl;
  return res;
}
signed main(){
  Int n,k;
  cin>>n>>k;
  
  po[0]=1;
  for(Int i=0;i<n;i++) po[i+1]=po[i]*(k+1)%MOD;
  
  memset(dp,-1,sizeof(dp));
  cout<<dfs(n,k,0)<<endl;
  return 0;
}

```