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

const int P=1000000007;
int n,k,dp[105][3282];

int power(int a,int x){
	int ans=1;
	for(;x;x>>=1,a=1LL*a*a%P)if(x&1)ans=1LL*ans*a%P;
	return ans;
}
void upd(int &x,int y){
	x=x+y>=P?x+y-P:x+y;
}

int main(){
	scanf("%d%d",&n,&k);
	dp[n+1][0]=1;
	for(int i=n;i;--i)
		for(int j=0;j<=3281;++j)
			if(dp[i+1][j]){
				for(int a=0;a<=i&&a<=k;++a)
					upd(dp[i][j+(a+j)/i],dp[i+1][j]);
				if(i<k)
					dp[i][j]=(dp[i][j]+1LL*(k-i)*dp[i+1][j])%P;
			}
	int ans=1LL*n*(1+k)*k/2*power(k+1,n-1)%P;
	for(int i=1;i<=3281;++i)
		ans=(ans-1LL*dp[1][i]*i)%P;
	printf("%d\n",(ans+P)%P);
}

### 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 int P=1000000007;
int n,k,dp[105][3282];

int power(int a,int x){
	int ans=1;
	for(;x;x>>=1,a=1LL*a*a%P)if(x&1)ans=1LL*ans*a%P;
	return ans;
}
void upd(int &x,int y){
	x=x+y>=P?x+y-P:x+y;
}

int main(){
	scanf("%d%d",&n,&k);
	dp[n+1][0]=1;
	for(int i=n;i;--i)
		for(int j=0;j<=3281;++j)
			if(dp[i+1][j]){
				for(int a=0;a<=i&&a<=k;++a)
					upd(dp[i][j+(a+j)/i],dp[i+1][j]);
				if(i<k)
					dp[i][j]=(dp[i][j]+1LL*(k-i)*dp[i+1][j])%P;
			}
	int ans=1LL*n*(1+k)*k/2*power(k+1,n-1)%P;
	for(int i=1;i<=3281;++i)
		ans=(ans-1LL*dp[1][i]*i)%P;
	printf("%d\n",(ans+P)%P);
}

```