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

### 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 int mod = 1000000007;
long long qpow(long long x, int a) {
  long long ret = 1;
  while (a) {
    if (a & 1) ret = ret * x % mod;
    x = x * x % mod;
    a >>= 1;
  }
  return ret;
}
long long n, m;
int main() {
  scanf("%lld %lld", &n, &m);
  long long t = qpow(2 * m - 1, n);
  long long ans =
      t + (((t - qpow(m, n) % mod + mod) % mod) * qpow(m - 1, mod - 2) % mod);
  if (m > 1)
    printf("%lld\n", ans % mod);
  else
    printf("%lld\n", n + 1);
  return 0;
}

```