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;
const int K = 5001;
const long long mod = 1e9 + 7;
int n, k;
long long h[K], rev[K], c[K][K], g[K], nf[K], res = 0;
long long hp(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x % mod;
  long long a = hp(x, y >> 1);
  a = (a * a) % mod;
  return y & 1 ? (a * x) % mod : a;
}
long long combination(int i, int n) {
  if (i < 0 || i > n) return 0;
  if (i == 0 || i == n) return 1;
  return rev[i] * nf[i] % mod;
}
long long StirlingNumberOfTheSecondKind(int i, int k) {
  long long rr = 0;
  for (int j = 0; j <= i; ++j)
    rr = (rr + c[j][i] * g[i - j] % mod * (j & 1 ? -1 : 1)) % mod;
  return rr * rev[i] % mod;
}
void testcase() {
  cin >> n >> k;
  h[0] = 1;
  for (int i = 1; i <= k; ++i) {
    h[i] = h[i - 1] * i % mod;
    rev[i] = hp(h[i], mod - 2);
  }
  nf[0] = 1;
  for (int i = 1; i <= k; ++i) nf[i] = nf[i - 1] * (n - i + 1) % mod;
  for (int i = 0; i <= k; ++i) {
    g[i] = hp(i, k);
  }
  for (int j = 0; j <= k; ++j) {
    for (int i = 0; i <= j; ++i) {
      if (i == 0 || i == j)
        c[i][j] = 1;
      else {
        c[i][j] = (c[i][j - 1] + c[i - 1][j - 1]) % mod;
      }
    }
  }
  for (int i = 1; i <= k && i <= n; ++i) {
    long long rr = combination(i, n) * hp(2, n - i) % mod;
    rr = rr * h[i] % mod * StirlingNumberOfTheSecondKind(i, k) % mod;
    res = (res + rr) % mod;
  }
  res = (res + mod) % mod;
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int tcase = 1;
  for (int i = 1; i <= tcase; ++i) testcase();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}