#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<string>
#include<vector>

using namespace std;

typedef long long ll;

const int MX = 105;
const int MM = 1000000007;

typedef pair<int,int> pii;

ll t[MX][MX*MX];

int add(int x, int y){
	return x+y >= MM? x+y-MM : x+y;
}

int main()
{
	int N, K;
	scanf("%d%d", &N, &K);
	t[N+1][0] = 1;
	for(int i = N; i >= 1; i--){
		for(int j = 0; j < MX*MX; j++){
			if( t[i+1][j] == 0 ) continue;
			for(int k = 0; k <= K; k++){
				if( k <= i ){
					int p = (k + j) / i;
					t[i][p + j] = add(t[i][p+j], t[i+1][j]);
				}
				else{
					t[i][j] = add(t[i][j], t[i+1][j]);
				}
			}
		}
	}
	ll m = 1;
	for(int i = 2; i <= N+1; i++){
		for(int j = 0; j < MX*MX; j++){
			t[i][j] = t[i][j] * m % MM;
		}
		if(i != N+1) m = m*(K+1) % MM;
	}
	ll ans = 0;
	for(int i = 1; i <= N; i++){
		ans = add(ans, K*(K+1)/2 * m % MM);
	}
	for(int i = 1; i <= N; i++){
		for(int j = 0; j < MX*MX; j++){
			if( t[i+1][j] == 0 ) continue;
			for(int k = 0; k <= K; k++){
				int p = (j + k) / i;
				if( k <= i ){
					ans = add(ans, MM - t[i+1][j] * p % MM);
				}
			}
		}
	}
	printf("%lld\n", ans);
}

### Prompt

Develop a solution in CPP 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<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<string>
#include<vector>

using namespace std;

typedef long long ll;

const int MX = 105;
const int MM = 1000000007;

typedef pair<int,int> pii;

ll t[MX][MX*MX];

int add(int x, int y){
	return x+y >= MM? x+y-MM : x+y;
}

int main()
{
	int N, K;
	scanf("%d%d", &N, &K);
	t[N+1][0] = 1;
	for(int i = N; i >= 1; i--){
		for(int j = 0; j < MX*MX; j++){
			if( t[i+1][j] == 0 ) continue;
			for(int k = 0; k <= K; k++){
				if( k <= i ){
					int p = (k + j) / i;
					t[i][p + j] = add(t[i][p+j], t[i+1][j]);
				}
				else{
					t[i][j] = add(t[i][j], t[i+1][j]);
				}
			}
		}
	}
	ll m = 1;
	for(int i = 2; i <= N+1; i++){
		for(int j = 0; j < MX*MX; j++){
			t[i][j] = t[i][j] * m % MM;
		}
		if(i != N+1) m = m*(K+1) % MM;
	}
	ll ans = 0;
	for(int i = 1; i <= N; i++){
		ans = add(ans, K*(K+1)/2 * m % MM);
	}
	for(int i = 1; i <= N; i++){
		for(int j = 0; j < MX*MX; j++){
			if( t[i+1][j] == 0 ) continue;
			for(int k = 0; k <= K; k++){
				int p = (j + k) / i;
				if( k <= i ){
					ans = add(ans, MM - t[i+1][j] * p % MM);
				}
			}
		}
	}
	printf("%lld\n", ans);
}

```