#include <bits/stdc++.h>
using namespace std;
int n, m;
long long po(long long a, long long b) {
  long long c = 1, d = a;
  while (b) {
    if (b & 1) c = 1LL * c * d % 1000000007;
    d = 1LL * d * d % 1000000007;
    b >>= 1;
  }
  return c;
}
int main() {
  scanf("%d %d", &n, &m);
  if (m == 1)
    printf("%d\n", n + 1);
  else
    printf("%lld\n",
           (((po(2 * m - 1, n) - po(m, n)) * po(m - 1, 1000000007 - 2) +
             po(2 * m - 1, n)) %
                1000000007 +
            1000000007) %
               1000000007);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
int n, m;
long long po(long long a, long long b) {
  long long c = 1, d = a;
  while (b) {
    if (b & 1) c = 1LL * c * d % 1000000007;
    d = 1LL * d * d % 1000000007;
    b >>= 1;
  }
  return c;
}
int main() {
  scanf("%d %d", &n, &m);
  if (m == 1)
    printf("%d\n", n + 1);
  else
    printf("%lld\n",
           (((po(2 * m - 1, n) - po(m, n)) * po(m - 1, 1000000007 - 2) +
             po(2 * m - 1, n)) %
                1000000007 +
            1000000007) %
               1000000007);
  return 0;
}

```