#include <bits/stdc++.h>
using namespace std;
long long f[120][4020], z;
int n, k, p = 1000000007;
int main() {
	cin >> n >> k;
	z = k * (k + 1) / 2 * n;
	for (int i = 1; i < n; i++) {
		z = z * (k + 1) % p;
	}
	f[n][0] = 1;
	for (int i = n; i > 0; i--) {
		for (int j = 0; j < 4010; j++) {
			if (f[i][j] == 0) {
				continue;
			}
			for (int l = 0; l <= k; l++) {
				if (l <= i) {
					f[i - 1][j + (j + l) / i] = (f[i - 1][j + (j + l) / i] + f[i][j]) % p;
				} else {
					f[i - 1][j] = (f[i - 1][j] + f[i][j]) % p;
				}
			}
		}
	}
	for (int i = 0; i < 4010; i++) {
		z = (z - i * f[0][i]) % p;
	}
	cout << (z + p) % p << endl;
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
long long f[120][4020], z;
int n, k, p = 1000000007;
int main() {
	cin >> n >> k;
	z = k * (k + 1) / 2 * n;
	for (int i = 1; i < n; i++) {
		z = z * (k + 1) % p;
	}
	f[n][0] = 1;
	for (int i = n; i > 0; i--) {
		for (int j = 0; j < 4010; j++) {
			if (f[i][j] == 0) {
				continue;
			}
			for (int l = 0; l <= k; l++) {
				if (l <= i) {
					f[i - 1][j + (j + l) / i] = (f[i - 1][j + (j + l) / i] + f[i][j]) % p;
				} else {
					f[i - 1][j] = (f[i - 1][j] + f[i][j]) % p;
				}
			}
		}
	}
	for (int i = 0; i < 4010; i++) {
		z = (z - i * f[0][i]) % p;
	}
	cout << (z + p) % p << endl;
	return 0;
}
```