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 PYTHON3):

Solution:
import sys


def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())


N,k = MI()
mod = 10**9+7

inv_2 = (mod+1)//2
x = pow(2,N,mod)
X = [0]*(k+1)  # X[i] = N_P_i*2^(N-i)
for i in range(k+1):
    X[i] = x
    x *= (N-i)*inv_2
    x %= mod

coefficient = [0]*(k+1)
# coefficient[i] = f(x)に対して、「微分して*xする」をi回繰り返した際の係数
# メモリ節約のため、i省略
coefficient[0] = 1
for _ in range(k):
    for j in range(k,-1,-1):
        coefficient[j] = coefficient[j]*j+coefficient[j-1]
        coefficient[j] %= mod

ans = 0
for i in range(1,k+1):
    ans += coefficient[i]*X[i]
    ans %= mod

print(ans)