#include<bits/stdc++.h>
using namespace std;
const long long M=1000000007;//can be changed
const int m=3282;
long long dp[101][m];

int main(){
	long long n,k;
	cin>>n>>k;
	long long ans=(n*(k*(k+1)/2))%M;
	for(int i=0;i<n-1;++i){
		ans*=(k+1);
		ans%=M;
	}
	fill(dp[n],dp[n]+m,0);
	dp[n][0]=1;
	for(int i=n;i>0;--i) for(int j=0;j<m;++j) for(int a=0;a<=k;++a){
		if(a<=i){
			dp[i-1][j+(a+j)/i]+=dp[i][j];
			dp[i-1][j+(a+j)/i]%=M;
		}
		else{
			dp[i-1][j]+=dp[i][j];
			dp[i-1][j]%=M;
		}
	}
	for(int j=0;j<m;++j) ans=(ans+M-(j*dp[0][j])%M)%M;
	cout<<ans<<endl;
}
### Prompt

Please create a solution in cpp 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;
const long long M=1000000007;//can be changed
const int m=3282;
long long dp[101][m];

int main(){
	long long n,k;
	cin>>n>>k;
	long long ans=(n*(k*(k+1)/2))%M;
	for(int i=0;i<n-1;++i){
		ans*=(k+1);
		ans%=M;
	}
	fill(dp[n],dp[n]+m,0);
	dp[n][0]=1;
	for(int i=n;i>0;--i) for(int j=0;j<m;++j) for(int a=0;a<=k;++a){
		if(a<=i){
			dp[i-1][j+(a+j)/i]+=dp[i][j];
			dp[i-1][j+(a+j)/i]%=M;
		}
		else{
			dp[i-1][j]+=dp[i][j];
			dp[i-1][j]%=M;
		}
	}
	for(int j=0;j<m;++j) ans=(ans+M-(j*dp[0][j])%M)%M;
	cout<<ans<<endl;
}
```