#include <bits/stdc++.h>
long long n, m, ans, tmp, i;
long long Pow(long long x, long long y) {
  if (!x) return 1;
  for (tmp = 1; y; y >>= 1, x = (x * x) % 1000000007)
    if (y & 1) tmp = tmp * x % 1000000007;
  return tmp;
}
int main() {
  scanf("%I64d%I64d", &n, &m);
  if (m == 1) {
    printf("%I64d", n + 1);
    return 0;
  }
  ans = Pow(m, n);
  long long R = Pow(m - 1, n), con = Pow(m - 1, 1000000007 - 2);
  for (i = 1; i <= n; i++)
    ans = (ans + (R = R * con % 1000000007 * m % 1000000007)) % 1000000007,
    R = R * ((n - i + 1) * Pow(i, 1000000007 - 2) % 1000000007) % 1000000007;
  printf("%I64d", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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, ans, tmp, i;
long long Pow(long long x, long long y) {
  if (!x) return 1;
  for (tmp = 1; y; y >>= 1, x = (x * x) % 1000000007)
    if (y & 1) tmp = tmp * x % 1000000007;
  return tmp;
}
int main() {
  scanf("%I64d%I64d", &n, &m);
  if (m == 1) {
    printf("%I64d", n + 1);
    return 0;
  }
  ans = Pow(m, n);
  long long R = Pow(m - 1, n), con = Pow(m - 1, 1000000007 - 2);
  for (i = 1; i <= n; i++)
    ans = (ans + (R = R * con % 1000000007 * m % 1000000007)) % 1000000007,
    R = R * ((n - i + 1) * Pow(i, 1000000007 - 2) % 1000000007) % 1000000007;
  printf("%I64d", ans);
  return 0;
}

```