#include <bits/stdc++.h>
using namespace std;
const long long P = 1000000007;
long long f1[1000010], f2[1000010], ans;
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  int i;
  for (f1[0] = f2[0] = i = 1; i <= n; i++)
    f1[i] = f1[i - 1] * m % P, f2[i] = f2[i - 1] * (m + m - 1) % P;
  for (ans = f1[n], i = 1; i <= n; i++)
    ans = (ans + f1[n - i + 1] * f2[i - 1]) % P;
  printf("%lld", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long P = 1000000007;
long long f1[1000010], f2[1000010], ans;
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  int i;
  for (f1[0] = f2[0] = i = 1; i <= n; i++)
    f1[i] = f1[i - 1] * m % P, f2[i] = f2[i - 1] * (m + m - 1) % P;
  for (ans = f1[n], i = 1; i <= n; i++)
    ans = (ans + f1[n - i + 1] * f2[i - 1]) % P;
  printf("%lld", ans);
  return 0;
}

```