#include <bits/stdc++.h>
using namespace std;
long long n,k;
long long ans=0,dp[107][10007],sub=0,sum=0;
const long long mod=1e9+7;
long long pw(long long a,long long b){
	if(b==0) return 1LL;
	if(b==1) return a;
	long long ret=pw(a,b/2);
	if(b%2==0) return ret*ret%mod;
	else return ret*ret%mod*a%mod;
}
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>k;
	ans=(n*(((k+1)*k)/2LL))%mod;
	ans=(ans*pw(k+1,n-1))%mod;
	dp[n+1][0]=1;
	for(int i=n;i>=1;i--){
		int sv=sum;
		for(int j=0;j<=sum;j++){
			for(int t=0;t<=k;t++){
				if(t<=i){
					dp[i][j+(j+t)/i]=(dp[i][j+(j+t)/i]+dp[i+1][j])%mod;
					sv=max(sv,j+j/i+1);
				}
				else{
					dp[i][j]=(dp[i][j]+dp[i+1][j])%mod;
				}
			}
		}
		while(dp[i][sv]==0) sv--;
		sum=sv;
	}
	for(int i=1;i<=sum;i++){
		sub=(sub+1LL*i*dp[1][i])%mod;
	}
	ans=ans-sub+mod;
	ans%=mod;
	cout<<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;
long long n,k;
long long ans=0,dp[107][10007],sub=0,sum=0;
const long long mod=1e9+7;
long long pw(long long a,long long b){
	if(b==0) return 1LL;
	if(b==1) return a;
	long long ret=pw(a,b/2);
	if(b%2==0) return ret*ret%mod;
	else return ret*ret%mod*a%mod;
}
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>k;
	ans=(n*(((k+1)*k)/2LL))%mod;
	ans=(ans*pw(k+1,n-1))%mod;
	dp[n+1][0]=1;
	for(int i=n;i>=1;i--){
		int sv=sum;
		for(int j=0;j<=sum;j++){
			for(int t=0;t<=k;t++){
				if(t<=i){
					dp[i][j+(j+t)/i]=(dp[i][j+(j+t)/i]+dp[i+1][j])%mod;
					sv=max(sv,j+j/i+1);
				}
				else{
					dp[i][j]=(dp[i][j]+dp[i+1][j])%mod;
				}
			}
		}
		while(dp[i][sv]==0) sv--;
		sum=sv;
	}
	for(int i=1;i<=sum;i++){
		sub=(sub+1LL*i*dp[1][i])%mod;
	}
	ans=ans-sub+mod;
	ans%=mod;
	cout<<ans;
}

```