"""

包除原理で解きそう
全部0で埋める行・列の個数によって変化してしまうな…

まず、行か列の片方だけ求める？
少なくともX個列が0で埋まっている時、少なくともY個行が0で埋まっている場合の数

O(HW)解は

Σ nCr(H,x) * nCr(W,y) * pow( 2 , H*W-x*W-H*y+x*y , mod ) * pow(-1,x+y,mod)
で求まる

分離できないか？

H*W - x*W - H*y + x*y
= (H-x)*(W-y)

分離できてないじゃん

採用   = 2^k
不採用 = -1

A[i+1] = A[i] * ( 2^k-1 )

"""

import sys
from sys import stdin

def modfac(n, MOD):
 
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs


def modnCr(n,r): 
    return fac[n] * inv[n-r] * inv[r] % mod

mod = 10**9+7
fac,inv = modfac(10**6+100,mod)
p2 = [1]
for i in range(10**6+100):
    p2.append(p2[-1]*2 % mod)

H,W = map(int,stdin.readline().split())

ans = 0
for unpick in range(0,H+1):

    pick = H-unpick
    bi = p2[pick]

    now = pow(bi-1,W,mod)
    ans += now * modnCr(H,unpick) * ( (-1)**(unpick%2) )
    ans %= mod

print (ans % mod)

"""
A = 0
for x in range(0,H+1):
    now = modnCr(H,x) * p2[H-x]
    if now % 2 == 0:
        A += now
    else:
        A -= now

B = 0
for y in range(0,W+1):
    now = modnCr(W,y) * p2[W-y]
    if now % 2 == 0:
        B += now
    else:
        B -= now

print (A,B)
print (A*B%mod)
"""