#include <bits/stdc++.h>
using namespace std;
long long ans, mod = 1000000007;
int n, m;
int mypow(int x, long long times) {
  if (times == 0) return 1;
  if (times == 1) return x;
  long long ret = mypow(x, times / 2);
  ret = ret * ret % mod;
  if (times % 2) ret = ret * x % mod;
  return ret;
}
int main() {
  cin >> n >> m;
  ans = mypow(m, n);
  for (int i = 0; i < n; i++)
    ans = (ans + 1ll * mypow(m, n - i) * mypow(2 * m - 1, i)) % mod;
  cout << ans;
}

### 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;
long long ans, mod = 1000000007;
int n, m;
int mypow(int x, long long times) {
  if (times == 0) return 1;
  if (times == 1) return x;
  long long ret = mypow(x, times / 2);
  ret = ret * ret % mod;
  if (times % 2) ret = ret * x % mod;
  return ret;
}
int main() {
  cin >> n >> m;
  ans = mypow(m, n);
  for (int i = 0; i < n; i++)
    ans = (ans + 1ll * mypow(m, n - i) * mypow(2 * m - 1, i)) % mod;
  cout << ans;
}

```