#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using mint = modint1000000007;
const int mod = 1000000007;
//using mint = modint998244353;
//const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) { return com(n, k); }
	mint com(int n, int k) {
		//負の二項係数を考慮する場合にコメントアウトを外す
		//if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
	mint inv(int n, int k) {
		//if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return ifact[n] * fact[k] * fact[n - k];
	}
	mint p(int n, int k) { return fact[n] * ifact[n - k]; }
}c(2000006);

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, k; cin >> n >> k;
	mint ans = pow_mod(n - 1, k, mod);
	int last = min(n - 1, k) + 1;
	int sz = max(0, n - last);
	rep2(i, 1, min(n - 1, k) + 2) {
		int j = i + 1;
		vector dp(k + 1, vector(i + 1, vector<mint>(j + 1, 0)));
		dp[0][i][j] = 1;
		rep(s, k) {
			rep(t, i + 1)rep(u, j + 1) {
				if (0 != t) {
					int nt = t - 1;
					int nu = u;
					dp[s + 1][nt][nu] += dp[s][t][u];
				}
				if ((t != (u - 1)) && (0 != u)) {
					int nt = t;
					int nu = u - 1;
					dp[s + 1][nt][nu] += dp[s][t][u];
				}
				int nt = t;
				int nu = u;
				int x = n;
				if (0 != t)x--;
				if (0 != u)x--;
				dp[s + 1][nt][nu] += dp[s][t][u] * x;
			}
		}
		rep(s, i + 1)rep(t, j + 1) {
			if (last != i)ans += dp[k][s][t] * t;
			else {
				int l = t;
				int r = t + sz - 1;
				ans += dp[k][s][t]  * (l + r) * (sz) / 2;
			}
		}
	}
	cout << ans.val() << endl;
	return 0;
}