#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long n, m;
long long ans;
long long pow(long long b, long long e) {
  long long ret = 1;
  while (e) {
    if (e & 1) ret = (ret * b) % mod;
    e >>= 1;
    b = (b * b) % mod;
  }
  return ret;
}
int main() {
  scanf("%I64d%I64d", &n, &m);
  if (m == 1) {
    ans = n + 1;
  } else {
    ans = ((m * pow(2 * m - 1, n) - pow(m, n + 1) + mod) % mod *
               pow(m - 1, mod - 2) +
           pow(m, n)) %
          mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long mod = 1000000007;
long long n, m;
long long ans;
long long pow(long long b, long long e) {
  long long ret = 1;
  while (e) {
    if (e & 1) ret = (ret * b) % mod;
    e >>= 1;
    b = (b * b) % mod;
  }
  return ret;
}
int main() {
  scanf("%I64d%I64d", &n, &m);
  if (m == 1) {
    ans = n + 1;
  } else {
    ans = ((m * pow(2 * m - 1, n) - pow(m, n + 1) + mod) % mod *
               pow(m - 1, mod - 2) +
           pow(m, n)) %
          mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

```