#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
const LL mod=1e9+7;
LL f[110][20010],n,k,tot;
LL Pow(LL a,LL b)
{
	LL ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld %lld",&n,&k);
	tot=Pow(k+1,n-1)*n%mod*k%mod*(k+1)%mod*((mod+1)/2)%mod;
	f[k+1][0]=1;
	for(LL i=k+1;i>1;i--)
		for(LL j=0;j<=20000;j++) if(f[i][j])
			for(LL u=0;u<=k;u++)
			{
				if(u>i-1) (f[i-1][j]+=f[i][j])%=mod;
				else (f[i-1][j+(j+u)/(i-1)]+=f[i][j])%=mod;
			}
	for(LL i=1;i<=20000;i++)
		(tot-=f[1][i]*i%mod*Pow(k+1,n-k)%mod)%=mod;
	(tot+=mod)%=mod;
	printf("%lld",tot);
}
### 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
const LL mod=1e9+7;
LL f[110][20010],n,k,tot;
LL Pow(LL a,LL b)
{
	LL ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld %lld",&n,&k);
	tot=Pow(k+1,n-1)*n%mod*k%mod*(k+1)%mod*((mod+1)/2)%mod;
	f[k+1][0]=1;
	for(LL i=k+1;i>1;i--)
		for(LL j=0;j<=20000;j++) if(f[i][j])
			for(LL u=0;u<=k;u++)
			{
				if(u>i-1) (f[i-1][j]+=f[i][j])%=mod;
				else (f[i-1][j+(j+u)/(i-1)]+=f[i][j])%=mod;
			}
	for(LL i=1;i<=20000;i++)
		(tot-=f[1][i]*i%mod*Pow(k+1,n-k)%mod)%=mod;
	(tot+=mod)%=mod;
	printf("%lld",tot);
}
```