#ifndef LOCAL
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const ll MOD = 1e9 + 7;
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}

const int MX = 6000;

int n, k;
ll dp[110][MX];

void add(ll &a, ll b) {
	a += b;
	if (a >= MOD)
		a -= MOD;
}

int main() {
	cin >> n >> k;
	dp[n + 1][0] = 1;
	for (int i = n; i >= 1; --i) {
		for (int j = 0; j < MX; ++j) {
			if (dp[i + 1][j] == 0)
				continue;
			for (int l = 0; l <= k; ++l) {
				if (l > i) {
					add(dp[i][j], dp[i + 1][j]);
				}
				else  {
					add(dp[i][j + (j + l) / i], dp[i + 1][j]);
				}
			}
		}
	}
	ll ans = n * pw(k + 1, n) % MOD * k % MOD * (MOD / 2 + 1) % MOD;
	for (int i = 0; i < MX; ++i)
		ans = (ans - i * dp[1][i] % MOD + MOD) % MOD;
	cout << ans << "\n";
	return 0;
}



### Prompt

Your task is to create 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
#ifndef LOCAL
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const ll MOD = 1e9 + 7;
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}

const int MX = 6000;

int n, k;
ll dp[110][MX];

void add(ll &a, ll b) {
	a += b;
	if (a >= MOD)
		a -= MOD;
}

int main() {
	cin >> n >> k;
	dp[n + 1][0] = 1;
	for (int i = n; i >= 1; --i) {
		for (int j = 0; j < MX; ++j) {
			if (dp[i + 1][j] == 0)
				continue;
			for (int l = 0; l <= k; ++l) {
				if (l > i) {
					add(dp[i][j], dp[i + 1][j]);
				}
				else  {
					add(dp[i][j + (j + l) / i], dp[i + 1][j]);
				}
			}
		}
	}
	ll ans = n * pw(k + 1, n) % MOD * k % MOD * (MOD / 2 + 1) % MOD;
	for (int i = 0; i < MX; ++i)
		ans = (ans - i * dp[1][i] % MOD + MOD) % MOD;
	cout << ans << "\n";
	return 0;
}



```