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.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
const long long MAXR = 5000;
long long s, inv[MAXR + 10], dp[MAXR + 2][MAXR + 2], fac[MAXR + 1];
long long nCr(long long n, long long r) {
  if (r > n) return 0;
  if (n - r < r) r = n - r;
  n %= mod;
  long long ans = 1;
  for (int i = 0; i < (int)r; i++) {
    ans = (ans * (n - i)) % mod;
    ans = (ans * inv[i + 1]) % mod;
  }
  return ans;
}
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
void pre() {
  fac[0] = 1;
  for (int i = 1; i <= MAXR; i++) {
    inv[i] = modpow(i, mod - 2);
    fac[i] = (i * fac[i - 1]) % mod;
  }
}
long long mul(vector<long long> nums) {
  for (int i = 1; i < (int)nums.size(); i++)
    nums[i] = (nums[i - 1] * nums[i]) % mod;
  return nums.back();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  pre();
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i < (int)k + 1; i++)
    for (int j = 1; j < (int)i + 1; j++) {
      if (j == i)
        dp[i][j] = 1;
      else {
        dp[i][j] = dp[i - 1][j - 1] + (j * dp[i - 1][j]) % mod;
        if (dp[i][j] >= mod) dp[i][j] -= mod;
      }
    }
  long long ans = 0;
  for (int d = 1; d < (int)min(n, k) + 1; d++) {
    ans += mul({nCr(n, d), modpow(2, n - d), dp[k][d], fac[d]});
    if (ans >= mod) ans -= mod;
  }
  cout << ans << '\n';
  return 0;
}
