#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n,k;
ll sum;
ll dp[101][3501];

int main(void){
	scanf("%d%d",&n,&k);
	sum=(ll)n*k*(k+1LL)/2LL;
	ll tmp=1;
	for(int i=0;i<n-1;i++){
		tmp=(ll)tmp*(k+1)%MOD;
	}
	sum=sum*tmp%MOD;
	dp[n][0]=1;
	for(int i=n-1;i>=0;i--){
		for(int j=0;j<=3400;j++){
			for(int l=0;l<=k;l++){
				if(l<=i+1){
					dp[i][j+(l+j)/(i+1)]+=dp[i+1][j];
					dp[i][j+(l+j)/(i+1)]%=MOD;
				}
				else{
					dp[i][j]+=dp[i+1][j];
					dp[i][j]%=MOD;
				}
			}
		}
	}
	for(ll i=1;i<3400;i++){
		sum-=dp[0][i]*i%MOD;
		if(sum<0LL)sum+=MOD;
	}
	printf("%lld\n",sum);
	return 0;
}
### Prompt

Please formulate a CPP solution 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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n,k;
ll sum;
ll dp[101][3501];

int main(void){
	scanf("%d%d",&n,&k);
	sum=(ll)n*k*(k+1LL)/2LL;
	ll tmp=1;
	for(int i=0;i<n-1;i++){
		tmp=(ll)tmp*(k+1)%MOD;
	}
	sum=sum*tmp%MOD;
	dp[n][0]=1;
	for(int i=n-1;i>=0;i--){
		for(int j=0;j<=3400;j++){
			for(int l=0;l<=k;l++){
				if(l<=i+1){
					dp[i][j+(l+j)/(i+1)]+=dp[i+1][j];
					dp[i][j+(l+j)/(i+1)]%=MOD;
				}
				else{
					dp[i][j]+=dp[i+1][j];
					dp[i][j]%=MOD;
				}
			}
		}
	}
	for(ll i=1;i<3400;i++){
		sum-=dp[0][i]*i%MOD;
		if(sum<0LL)sum+=MOD;
	}
	printf("%lld\n",sum);
	return 0;
}
```