#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", n + 1);
    return 0;
  }
  int ans = 1LL * m * (power(2 * m - 1, n, mod) - power(m, n, mod) + mod) % mod;
  ans = 1LL * ans * power(m - 1, mod - 2, mod) % mod;
  ans = (ans + power(m, n, mod)) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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>
using namespace std;
const int mod = 1e9 + 7;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", n + 1);
    return 0;
  }
  int ans = 1LL * m * (power(2 * m - 1, n, mod) - power(m, n, mod) + mod) % mod;
  ans = 1LL * ans * power(m - 1, mod - 2, mod) % mod;
  ans = (ans + power(m, n, mod)) % mod;
  printf("%d\n", ans);
  return 0;
}

```