#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 7;
int n, m, i;
long long p, coeff;
long long f[2000020];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  coeff = 2 * m - 1;
  f[0] = p = 1ll;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1] * coeff;
    f[i] %= mod;
    f[i] += p;
    f[i] %= mod;
    p = (m * p) % mod;
  }
  cout << f[n];
  return 0;
}

### Prompt

Generate a cpp solution 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 mod = (long long)1e9 + 7;
int n, m, i;
long long p, coeff;
long long f[2000020];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  coeff = 2 * m - 1;
  f[0] = p = 1ll;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1] * coeff;
    f[i] %= mod;
    f[i] += p;
    f[i] %= mod;
    p = (m * p) % mod;
  }
  cout << f[n];
  return 0;
}

```