#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long modpow(long long i, long long j) {
  long long ans = 1;
  for (; j; j >>= 1, (i *= i) %= mod)
    if (j & 1) (ans *= i) %= mod;
  return ans;
}
int main() {
  long long n, m;
  cin >> n >> m;
  if (m == 1) return cout << n + 1, 0;
  cout << (modpow(m, n) +
           m * modpow(m - 1, mod - 2) % mod *
               (modpow(2 * m - 1, n) - modpow(m, n)) % mod +
           2 * mod) %
              mod;
}

### 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 modpow(long long i, long long j) {
  long long ans = 1;
  for (; j; j >>= 1, (i *= i) %= mod)
    if (j & 1) (ans *= i) %= mod;
  return ans;
}
int main() {
  long long n, m;
  cin >> n >> m;
  if (m == 1) return cout << n + 1, 0;
  cout << (modpow(m, n) +
           m * modpow(m - 1, mod - 2) % mod *
               (modpow(2 * m - 1, n) - modpow(m, n)) % mod +
           2 * mod) %
              mod;
}

```