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

```