#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 105;
const int MAXM = 3505;
const int mod = 1e9 + 7;

int n, m, ret, lst, f[MAXN][MAXM];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read(), m = Read();
	ret = n * m * (m + 1) / 2;
	for (int i = 1; i < n; i ++)
		ret = 1LL * ret * (m + 1) % mod;
	f[n + 1][0] = 1;
	for (int i = n; i; lst += (lst + min(i, m)) / i, i --)
		for (int j = 0; j <= lst; j ++)
			for (int k = 0; k <= m; k ++)
				if (k <= i)
					f[i][j + (j + k) / i] = (f[i][j + (j + k) / i] + f[i + 1][j]) % mod;
				else
					f[i][j] = (f[i][j] + f[i + 1][j]) % mod;
	for (int i = 1; i <= lst; i ++)
		ret = (ret - 1LL * f[1][i] * i % mod + mod) % mod;
	return printf("%d\n", ret), 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 105;
const int MAXM = 3505;
const int mod = 1e9 + 7;

int n, m, ret, lst, f[MAXN][MAXM];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read(), m = Read();
	ret = n * m * (m + 1) / 2;
	for (int i = 1; i < n; i ++)
		ret = 1LL * ret * (m + 1) % mod;
	f[n + 1][0] = 1;
	for (int i = n; i; lst += (lst + min(i, m)) / i, i --)
		for (int j = 0; j <= lst; j ++)
			for (int k = 0; k <= m; k ++)
				if (k <= i)
					f[i][j + (j + k) / i] = (f[i][j + (j + k) / i] + f[i + 1][j]) % mod;
				else
					f[i][j] = (f[i][j] + f[i + 1][j]) % mod;
	for (int i = 1; i <= lst; i ++)
		ret = (ret - 1LL * f[1][i] * i % mod + mod) % mod;
	return printf("%d\n", ret), 0;
}

```