Problem: You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk. 

Output the sum of costs over all non-empty subsets of people.

Input

Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).

Output

Output the sum of costs for all non empty subsets modulo 109 + 7.

Examples

Input

1 1


Output

1


Input

3 2


Output

24

Note

In the first example, there is only one non-empty subset {1} with cost 11 = 1.

In the second example, there are seven non-empty subsets.

- {1} with cost 12 = 1

- {2} with cost 12 = 1

- {1, 2} with cost 22 = 4

- {3} with cost 12 = 1

- {1, 3} with cost 22 = 4

- {2, 3} with cost 22 = 4

- {1, 2, 3} with cost 32 = 9

The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
long long modpow(long long a, long long b, long long m) {
  long long p = 1, q = a;
  for (int i = 0; i < 31; i++) {
    if ((b / (1LL << i)) % 2 == 1) {
      p *= q;
      p %= m;
    }
    q *= q;
    q %= m;
  }
  return p;
}
long long mod = 1000000007;
long long N, K, fact[5009], factinv[5009], modpows[5009];
void init() {
  fact[0] = 1;
  for (int i = 1; i <= 5006; i++) fact[i] = (1LL * fact[i - 1] * i) % mod;
  for (int i = 0; i <= 5006; i++) factinv[i] = modpow(fact[i], mod - 2, mod);
}
long long ncr(long long n, long long r) {
  long long e = 1;
  for (long long i = n; i >= n - r + 1; i--) {
    e *= i;
    e %= mod;
  }
  return e * factinv[r] % mod;
}
long long Lagrange(long long pos, vector<pair<long long, long long>> vec) {
  long long sum = 0;
  for (int i = 0; i < vec.size(); i++) {
    long long v1 = 1, v2 = 1;
    for (int j = 0; j < vec.size(); j++) {
      if (i == j) continue;
      v1 *= (pos - vec[j].first + mod) % mod;
      v1 %= mod;
      v2 *= (vec[i].first - vec[j].first + mod) % mod;
      v2 %= mod;
    }
    long long v = v1 * modpow(v2, mod - 2, mod) % mod;
    v *= vec[i].second;
    v %= mod;
    sum += v;
    sum %= mod;
  }
  return sum;
}
int main() {
  init();
  cin >> N >> K;
  for (int i = 0; i <= 5006; i++) modpows[i] = modpow(i, K, mod);
  if (N < K + 1) {
    long long V = 0;
    for (int i = 1; i <= N; i++) {
      V += ncr(N, i) * modpow(i, K, mod);
      V %= mod;
    }
    cout << V << endl;
  } else {
    vector<pair<long long, long long>> vec;
    for (int t = 1; t <= K + 1; t++) {
      long long s = 0, v = 1;
      for (int i = 1; i <= t; i++) {
        v *= (t + 1LL - 1LL * i);
        v %= mod;
        s += (v * factinv[i] % mod) * modpows[i];
        s %= mod;
      }
      s *= modpow(modpow(2, t, mod), mod - 2, mod);
      s %= mod;
      vec.push_back(make_pair(t, s));
    }
    long long e = Lagrange(N, vec);
    e *= modpow(2, N, mod);
    e %= mod;
    cout << e << endl;
  }
  return 0;
}