#include <bits/stdc++.h>
long long n, m;
long long ans;
long long ksm(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) ans = ans * a % p;
  return ans % p;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  if (m == 1) return 0 & printf("%lld", n + 1);
  ans = (ksm(m, n, 1000000007) +
         ((ksm(2 * m - 1, n, 1000000007) - ksm(m, n, 1000000007)) % 1000000007 +
          1000000007) *
             m % 1000000007 * ksm(m - 1, 1000000007 - 2, 1000000007) %
             1000000007) %
        1000000007;
  printf("%lld", ans);
}

### Prompt

Please formulate a CPP solution to the following problem:

For a sequence a of n integers between 1 and m, inclusive, denote f(a) as the number of distinct subsequences of a (including the empty subsequence).

You are given two positive integers n and m. Let S be the set of all sequences of length n consisting of numbers from 1 to m. Compute the sum f(a) over all a in S modulo 109 + 7.

Input

The only line contains two integers n and m (1 ≤ n, m ≤ 106) — the number of elements in arrays and the upper bound for elements.

Output

Print the only integer c — the desired sum modulo 109 + 7.

Examples

Input

1 3


Output

6


Input

2 2


Output

14


Input

3 3


Output

174

### Response

```cpp
#include <bits/stdc++.h>
long long n, m;
long long ans;
long long ksm(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) ans = ans * a % p;
  return ans % p;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  if (m == 1) return 0 & printf("%lld", n + 1);
  ans = (ksm(m, n, 1000000007) +
         ((ksm(2 * m - 1, n, 1000000007) - ksm(m, n, 1000000007)) % 1000000007 +
          1000000007) *
             m % 1000000007 * ksm(m - 1, 1000000007 - 2, 1000000007) %
             1000000007) %
        1000000007;
  printf("%lld", ans);
}

```