#include <iostream>
using namespace std;
long long dp[109][5509], N, K, mod = 1000000007;
int main() {
	cin >> N >> K; dp[N][0] = 1;
	for (int i = N; i >= 1; i--) {
		for (int j = 0; j <= N*(N + 1) / 2; j++) {
			if (dp[i][j] == 0)continue;
			for (int k = 0; k <= K; k++) {
				if (i < k) {
					dp[i - 1][j] += dp[i][j];
					dp[i - 1][j] %= mod;
				}
				else {
					int Z = j + k; Z /= i;
					dp[i - 1][j + Z] += dp[i][j];
					dp[i - 1][j + Z] %= mod;
				}
			}
		}
	}
	long long R = 1;
	for (int i = 0; i < N; i++) { R *= (K + 1); R %= mod; }
	R *= K*N; R %= mod; if (R % 2 == 1)R += mod; R /= 2;
	for (int i = 0; i <= N*(N + 1) / 2; i++) {
		R -= dp[0][i] * i; R = (R + mod*mod) % mod;
	}
	cout << R << endl;
	return 0;
}
### Prompt

Please create a solution in cpp 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 <iostream>
using namespace std;
long long dp[109][5509], N, K, mod = 1000000007;
int main() {
	cin >> N >> K; dp[N][0] = 1;
	for (int i = N; i >= 1; i--) {
		for (int j = 0; j <= N*(N + 1) / 2; j++) {
			if (dp[i][j] == 0)continue;
			for (int k = 0; k <= K; k++) {
				if (i < k) {
					dp[i - 1][j] += dp[i][j];
					dp[i - 1][j] %= mod;
				}
				else {
					int Z = j + k; Z /= i;
					dp[i - 1][j + Z] += dp[i][j];
					dp[i - 1][j + Z] %= mod;
				}
			}
		}
	}
	long long R = 1;
	for (int i = 0; i < N; i++) { R *= (K + 1); R %= mod; }
	R *= K*N; R %= mod; if (R % 2 == 1)R += mod; R /= 2;
	for (int i = 0; i <= N*(N + 1) / 2; i++) {
		R -= dp[0][i] * i; R = (R + mod*mod) % mod;
	}
	cout << R << endl;
	return 0;
}
```