/*input
20 17
*/
#include <bits/stdc++.h>
using namespace std;

const int N = 105;
const int mod = 1e9 + 7;

int f[N][N << 6];
int g[N][N << 6];

void upd(int &a, int b) {
	a = (a + b) % mod;
}

int main(){
	int n, k; cin >> n >> k;

	g[n + 1][0] = 1;	
	for(int i = n + 1; i > 1; -- i)
		for(int j = 0; j < 6000; ++ j) if(g[i][j])
			for(int cur = 0; cur <= k; ++ cur) {
				if(cur > i - 1) {
					upd(g[i - 1][j], g[i][j]);
					upd(f[i - 1][j], (f[i][j] + 1ll * (cur + j) * g[i][j]) % mod);
				}
				else {
					int k = (cur + j) / (i - 1);
					upd(g[i - 1][j + k], g[i][j]);
					upd(f[i - 1][j + k], (f[i][j] + 1ll * ((cur + j) % (i - 1)) * g[i][j]) % mod);
				}
			}

	cout << accumulate(f[1], f[1] + 6000, 0ll) % mod << '\n';
}

### Prompt

Generate 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
/*input
20 17
*/
#include <bits/stdc++.h>
using namespace std;

const int N = 105;
const int mod = 1e9 + 7;

int f[N][N << 6];
int g[N][N << 6];

void upd(int &a, int b) {
	a = (a + b) % mod;
}

int main(){
	int n, k; cin >> n >> k;

	g[n + 1][0] = 1;	
	for(int i = n + 1; i > 1; -- i)
		for(int j = 0; j < 6000; ++ j) if(g[i][j])
			for(int cur = 0; cur <= k; ++ cur) {
				if(cur > i - 1) {
					upd(g[i - 1][j], g[i][j]);
					upd(f[i - 1][j], (f[i][j] + 1ll * (cur + j) * g[i][j]) % mod);
				}
				else {
					int k = (cur + j) / (i - 1);
					upd(g[i - 1][j + k], g[i][j]);
					upd(f[i - 1][j + k], (f[i][j] + 1ll * ((cur + j) % (i - 1)) * g[i][j]) % mod);
				}
			}

	cout << accumulate(f[1], f[1] + 6000, 0ll) % mod << '\n';
}

```