#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
ll dp1[102][20001];
ll dp2[102][20001];
int main()
{
	int num, kai;
	scanf("%d%d", &num, &kai);
	dp1[num + 1][0] = 1;
	dp2[num + 1][0] = 0;
	int maxi = 0;
	for (int i = num; i >= 1; i--)
	{
		int xx = 0;
		for (int j = 0; j <= maxi; j++)
		{
			for (int k = 0; k <= kai; k++)
			{
				if (k > i)dp1[i][j] = (dp1[i][j] + dp1[i + 1][j]) % mod, dp2[i][j] = (dp2[i][j] + dp2[i + 1][j] + dp1[i + 1][j] * (j + k)) % mod;
				else dp1[i][(j + (j + k) / i)] = (dp1[i][(j + (j + k) / i)] + dp1[i + 1][j]) % mod, dp2[i][j + (j + k) / i] = (dp2[i][(j + (j + k) / i)] + dp2[i + 1][j] + dp1[i + 1][j] * ((j + k) % i)) % mod;
				xx = max(xx, j + (j + k) / i);
			}
		}
		maxi = xx;
	}
	ll sum = 0;
	for (int i = 0; i <= maxi; i++)sum += dp2[1][i];
	printf("%lld\n", sum%mod);
}
### 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
ll dp1[102][20001];
ll dp2[102][20001];
int main()
{
	int num, kai;
	scanf("%d%d", &num, &kai);
	dp1[num + 1][0] = 1;
	dp2[num + 1][0] = 0;
	int maxi = 0;
	for (int i = num; i >= 1; i--)
	{
		int xx = 0;
		for (int j = 0; j <= maxi; j++)
		{
			for (int k = 0; k <= kai; k++)
			{
				if (k > i)dp1[i][j] = (dp1[i][j] + dp1[i + 1][j]) % mod, dp2[i][j] = (dp2[i][j] + dp2[i + 1][j] + dp1[i + 1][j] * (j + k)) % mod;
				else dp1[i][(j + (j + k) / i)] = (dp1[i][(j + (j + k) / i)] + dp1[i + 1][j]) % mod, dp2[i][j + (j + k) / i] = (dp2[i][(j + (j + k) / i)] + dp2[i + 1][j] + dp1[i + 1][j] * ((j + k) % i)) % mod;
				xx = max(xx, j + (j + k) / i);
			}
		}
		maxi = xx;
	}
	ll sum = 0;
	for (int i = 0; i <= maxi; i++)sum += dp2[1][i];
	printf("%lld\n", sum%mod);
}
```