#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

const int mod = 1000000007;

long long dp[101][4000];
int main() {
	int n, k;
	scanf("%d %d", &n, &k);
	long long ans = (n * (k * (k + 1) / 2)) % mod;
	for (int i = 0; i < n - 1; ++ i)
		(ans *= (k + 1)) %= mod;
	for (int i = 0; i < 4000; ++ i)
		dp[n][i] = 0;
	dp[n][0] = 1;
	for (int i = n; i > 0; -- i)
		for (int j = 0; j < 4000; ++ j)
			for (int a = 0; a <= k; ++ a)
				if (a <= i)
					(dp[i - 1][j + (a + j) / i] += dp[i][j]) %= mod;
				else
					(dp[i - 1][j] += dp[i][j]) %= mod;
	for (int j = 0; j < 4000; ++ j)
		ans = (ans + mod - j * dp[0][j] % mod) % mod;
	printf("%lld\n", ans);
	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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

const int mod = 1000000007;

long long dp[101][4000];
int main() {
	int n, k;
	scanf("%d %d", &n, &k);
	long long ans = (n * (k * (k + 1) / 2)) % mod;
	for (int i = 0; i < n - 1; ++ i)
		(ans *= (k + 1)) %= mod;
	for (int i = 0; i < 4000; ++ i)
		dp[n][i] = 0;
	dp[n][0] = 1;
	for (int i = n; i > 0; -- i)
		for (int j = 0; j < 4000; ++ j)
			for (int a = 0; a <= k; ++ a)
				if (a <= i)
					(dp[i - 1][j + (a + j) / i] += dp[i][j]) %= mod;
				else
					(dp[i - 1][j] += dp[i][j]) %= mod;
	for (int j = 0; j < 4000; ++ j)
		ans = (ans + mod - j * dp[0][j] % mod) % mod;
	printf("%lld\n", ans);
	return 0;
}
```