#include <iostream>
#define int long long
#define MOD 1000000007
using namespace std;

int n,k,dp[110][110][2010],sum[110][2010];

signed main()
{
    cin>>n>>k;
    int p=1;
    for(int i=1;i<=n;i++){
    	for(int j=0;j<=k;j++){
    		for(int l=0;l<=2000;l++){
    			if(j>i){
    				dp[i][j][l]=sum[i-1][l]+(((l+j)*p)%MOD);
    				dp[i][j][l]%=MOD;
				}
				else{
					int val=j+l;
					int num=val/i,rem=val%i;
					if(l+num<=2000){
    					dp[i][j][l]=sum[i-1][l+num]+((rem*p)%MOD);
    					dp[i][j][l]%=MOD;
    				}
				}
    		}
    		for(int l=0;l<=2000;l++){
    			sum[i][l]+=dp[i][j][l];
    			sum[i][l]%=MOD;
    		}
    	}
    	p*=(k+1);
    	p%=MOD;
    }
    cout<<sum[n][0]<<endl;
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#define int long long
#define MOD 1000000007
using namespace std;

int n,k,dp[110][110][2010],sum[110][2010];

signed main()
{
    cin>>n>>k;
    int p=1;
    for(int i=1;i<=n;i++){
    	for(int j=0;j<=k;j++){
    		for(int l=0;l<=2000;l++){
    			if(j>i){
    				dp[i][j][l]=sum[i-1][l]+(((l+j)*p)%MOD);
    				dp[i][j][l]%=MOD;
				}
				else{
					int val=j+l;
					int num=val/i,rem=val%i;
					if(l+num<=2000){
    					dp[i][j][l]=sum[i-1][l+num]+((rem*p)%MOD);
    					dp[i][j][l]%=MOD;
    				}
				}
    		}
    		for(int l=0;l<=2000;l++){
    			sum[i][l]+=dp[i][j][l];
    			sum[i][l]%=MOD;
    		}
    	}
    	p*=(k+1);
    	p%=MOD;
    }
    cout<<sum[n][0]<<endl;
	return 0;
}
```