#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=105,M=1e9+7;
int n,m,dp[N][N*N],ans;
signed main(){
	scanf("%lld%lld",&n,&m);
	dp[n][0]++;
	for (int i=n;i;i--)
		for (int j=0;j<=(n-i+1)*m;j++)
			if (dp[i][j])for (int k=0;k<=m;k++)
				if (k<=i)(dp[i-1][j+(k+j)/i]+=dp[i][j])%=M;
				else (dp[i-1][j]+=dp[i][j])%=M;
	for (int j=0;j<=n*m;j++)(ans+=M-dp[0][j]*j%M)%=M;
	int fac=1;		
	for (int i=1;i<n;i++)(fac*=m+1)%=M;
	for (int i=0;i<=m;i++)(ans+=fac*n%M*i)%=M;
	printf("%lld\n",ans);
}
### Prompt

Create a solution in CPP for 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
const int N=105,M=1e9+7;
int n,m,dp[N][N*N],ans;
signed main(){
	scanf("%lld%lld",&n,&m);
	dp[n][0]++;
	for (int i=n;i;i--)
		for (int j=0;j<=(n-i+1)*m;j++)
			if (dp[i][j])for (int k=0;k<=m;k++)
				if (k<=i)(dp[i-1][j+(k+j)/i]+=dp[i][j])%=M;
				else (dp[i-1][j]+=dp[i][j])%=M;
	for (int j=0;j<=n*m;j++)(ans+=M-dp[0][j]*j%M)%=M;
	int fac=1;		
	for (int i=1;i<n;i++)(fac*=m+1)%=M;
	for (int i=0;i<=m;i++)(ans+=fac*n%M*i)%=M;
	printf("%lld\n",ans);
}
```