#include <bits/stdc++.h>
using namespace std;
const int p = 1e9 + 7;
long long fast_pow(long long x, long long y) {
  long long ans = 1, now = x;
  while (y) {
    if (y & 1) ans = ans * now % p;
    now = now * now % p;
    y >>= 1;
  }
  return ans;
}
int n, m;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    (ans += fast_pow(m, n + 1 - i) * fast_pow(2 * m - 1, i - 1)) %= p;
  printf("%lld", (ans + fast_pow(m, n)) % p);
  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;
const int p = 1e9 + 7;
long long fast_pow(long long x, long long y) {
  long long ans = 1, now = x;
  while (y) {
    if (y & 1) ans = ans * now % p;
    now = now * now % p;
    y >>= 1;
  }
  return ans;
}
int n, m;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    (ans += fast_pow(m, n + 1 - i) * fast_pow(2 * m - 1, i - 1)) %= p;
  printf("%lld", (ans + fast_pow(m, n)) % p);
  return 0;
}

```