#include <bits/stdc++.h>

constexpr int N = 100, P = 1000000007;

int n, k;
int dp[N + 1][N * N + 1];

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    std::cin >> n >> k;
    
    dp[n][0] = 1;
    for (int i = n; i > 0; --i) {
        for (int j = 0; j <= n * n; ++j) {
            for (int v = 0; v <= k; ++v) {
                if (dp[i][j] == 0)
                    continue;
                int t = v <= i ? (v + j) / i : 0;
                dp[i - 1][j + t] = (dp[i - 1][j + t] + dp[i][j]) % P;
            }
        }
    }
    
    int ans = 0;
    for (int i = 0; i <= n * n; ++i)
        ans = (ans + 1ll * i * dp[0][i]) % P;
    int tot = 1ll * k * n * (P + 1) / 2 % P;
    for (int i = 0; i < n; ++i)
        tot = 1ll * tot * (k + 1) % P;
    ans = (tot - ans + P) % P;
    std::cout << ans << "\n";
    
    return 0;
}
### Prompt

In CPP, your task is to solve 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 <bits/stdc++.h>

constexpr int N = 100, P = 1000000007;

int n, k;
int dp[N + 1][N * N + 1];

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    std::cin >> n >> k;
    
    dp[n][0] = 1;
    for (int i = n; i > 0; --i) {
        for (int j = 0; j <= n * n; ++j) {
            for (int v = 0; v <= k; ++v) {
                if (dp[i][j] == 0)
                    continue;
                int t = v <= i ? (v + j) / i : 0;
                dp[i - 1][j + t] = (dp[i - 1][j + t] + dp[i][j]) % P;
            }
        }
    }
    
    int ans = 0;
    for (int i = 0; i <= n * n; ++i)
        ans = (ans + 1ll * i * dp[0][i]) % P;
    int tot = 1ll * k * n * (P + 1) / 2 % P;
    for (int i = 0; i < n; ++i)
        tot = 1ll * tot * (k + 1) % P;
    ans = (tot - ans + P) % P;
    std::cout << ans << "\n";
    
    return 0;
}
```