#include <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
#define repeat_reverse(i, n) for (int i = (n)-1; (i) >= 0; --(i))
using ll = long long;
using namespace std;

constexpr int mod = 1e9+7;
int main() {
    int n, k; scanf("%d%d", &n, &k);
    const int dp_size = n * n + 3;
    vector<ll> cur(dp_size);
    cur[0] = 1;
    ll result = 0;
    repeat_reverse (i, n) {
        vector<ll> prv = move(cur);
        cur.assign(dp_size, 0);
        ll nxt = 1; repeat (j, i) nxt = nxt * (k + 1) % mod;
        repeat (a_i, min(k, i + 1) + 1) {
            repeat (delta, dp_size) if (prv[delta]) {
                result += (a_i + delta) % (i + 1) * nxt % mod * prv[delta] % mod;
                cur[delta + (a_i + delta) / (i + 1)] += prv[delta];
            }
        }
        repeat_from (a_i, (i + 1) + 1, k + 1) {
            repeat (delta, dp_size) if (prv[delta]) {
                result += (a_i + delta) * nxt % mod * prv[delta] % mod;
                cur[delta] += prv[delta];
            }
        }
        repeat (delta, dp_size) if (cur[delta]) {
            cur[delta] %= mod;
        }
        result %= mod;
    }
    printf("%lld\n", result);
    return 0;
}

### 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
#include <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
#define repeat_reverse(i, n) for (int i = (n)-1; (i) >= 0; --(i))
using ll = long long;
using namespace std;

constexpr int mod = 1e9+7;
int main() {
    int n, k; scanf("%d%d", &n, &k);
    const int dp_size = n * n + 3;
    vector<ll> cur(dp_size);
    cur[0] = 1;
    ll result = 0;
    repeat_reverse (i, n) {
        vector<ll> prv = move(cur);
        cur.assign(dp_size, 0);
        ll nxt = 1; repeat (j, i) nxt = nxt * (k + 1) % mod;
        repeat (a_i, min(k, i + 1) + 1) {
            repeat (delta, dp_size) if (prv[delta]) {
                result += (a_i + delta) % (i + 1) * nxt % mod * prv[delta] % mod;
                cur[delta + (a_i + delta) / (i + 1)] += prv[delta];
            }
        }
        repeat_from (a_i, (i + 1) + 1, k + 1) {
            repeat (delta, dp_size) if (prv[delta]) {
                result += (a_i + delta) * nxt % mod * prv[delta] % mod;
                cur[delta] += prv[delta];
            }
        }
        repeat (delta, dp_size) if (cur[delta]) {
            cur[delta] %= mod;
        }
        result %= mod;
    }
    printf("%lld\n", result);
    return 0;
}

```