#ifndef ___CLASS_MODINT
#define ___CLASS_MODINT

#include <cstdint>

template<std::uint32_t mod>
class modint {
private:
	std::uint32_t n;
public:
	modint() : n(0) {};
	modint(std::uint64_t n_) : n(n_ % mod) {};
	bool operator==(const modint& m) const { return n == m.n; }
	bool operator!=(const modint& m) const { return n != m.n; }
	std::uint32_t get() const { return n; }
	modint& operator+=(const modint& m) { n += m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator-=(const modint& m) { n += mod - m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator*=(const modint& m) { n = std::uint64_t(n) * m.n % mod; return *this; }
	modint operator+(const modint& m) const { return modint(*this) += m; }
	modint operator-(const modint& m) const { return modint(*this) -= m; }
	modint operator*(const modint& m) const { return modint(*this) *= m; }
	modint binpow(std::uint64_t b) const {
		modint ans = 1, m = modint(*this);
		while (b) {
			if (b & 1) ans *= m;
			m *= m;
			b >>= 1;
		}
		return ans;
	}
	modint inv() { return (*this).binpow(mod - 2); }
};

#endif // ___CLASS_MODINT

#include <vector>
#include <iostream>
using namespace std;
using modulo = modint<1000000007>;
int main() {
	int N, K;
	cin >> N >> K;
	vector<modulo> dp = { 1 };
	int sum = 0;
	for (int i = N; i >= 1; --i) {
		sum += (sum + i) / i;
		vector<modulo> ndp(sum + 1);
		for (int j = 0; j < dp.size(); ++j) {
			for (int k = 0; k <= K; ++k) {
				ndp[j + (k <= i ? (j + k) / i : 0)] += dp[j];
			}
		}
		dp = ndp;
	}
	modulo ans = 0;
	for (int i = 0; i <= sum; ++i) {
		ans += dp[i] * i;
	}
	modulo base = modulo(K + 1).binpow(N) * N * K * modulo(2).inv();
	cout << (base - ans).get() << endl;
	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
#ifndef ___CLASS_MODINT
#define ___CLASS_MODINT

#include <cstdint>

template<std::uint32_t mod>
class modint {
private:
	std::uint32_t n;
public:
	modint() : n(0) {};
	modint(std::uint64_t n_) : n(n_ % mod) {};
	bool operator==(const modint& m) const { return n == m.n; }
	bool operator!=(const modint& m) const { return n != m.n; }
	std::uint32_t get() const { return n; }
	modint& operator+=(const modint& m) { n += m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator-=(const modint& m) { n += mod - m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator*=(const modint& m) { n = std::uint64_t(n) * m.n % mod; return *this; }
	modint operator+(const modint& m) const { return modint(*this) += m; }
	modint operator-(const modint& m) const { return modint(*this) -= m; }
	modint operator*(const modint& m) const { return modint(*this) *= m; }
	modint binpow(std::uint64_t b) const {
		modint ans = 1, m = modint(*this);
		while (b) {
			if (b & 1) ans *= m;
			m *= m;
			b >>= 1;
		}
		return ans;
	}
	modint inv() { return (*this).binpow(mod - 2); }
};

#endif // ___CLASS_MODINT

#include <vector>
#include <iostream>
using namespace std;
using modulo = modint<1000000007>;
int main() {
	int N, K;
	cin >> N >> K;
	vector<modulo> dp = { 1 };
	int sum = 0;
	for (int i = N; i >= 1; --i) {
		sum += (sum + i) / i;
		vector<modulo> ndp(sum + 1);
		for (int j = 0; j < dp.size(); ++j) {
			for (int k = 0; k <= K; ++k) {
				ndp[j + (k <= i ? (j + k) / i : 0)] += dp[j];
			}
		}
		dp = ndp;
	}
	modulo ans = 0;
	for (int i = 0; i <= sum; ++i) {
		ans += dp[i] * i;
	}
	modulo base = modulo(K + 1).binpow(N) * N * K * modulo(2).inv();
	cout << (base - ans).get() << endl;
	return 0;
}
```