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

### Prompt

Develop a solution in cpp to the problem described below:

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 add(int a, int b) { return a + b - (a + b >= MOD) * MOD; }
int sub(int a, int b) { return a - b + (a - b < 0) * MOD; }
int mult(int a, int b) { return (1LL * a * b) % MOD; }
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  int pow = 1;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    ans = mult(ans, 2 * m - 1);
    ans = sub(ans, mult(m - 1, pow));
    pow = mult(pow, m);
    ans = add(ans, pow);
  }
  printf("%d\n", ans);
  return 0;
}

```