#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e6 + 4;
long long me(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return res;
}
long long mi(long long n) { return me(n, mod - 2); }
int main() {
  long long n, m;
  scanf("%lld", &n);
  scanf("%lld", &m);
  if (m == 1) {
    printf("%lld", n + 1);
    return 0;
  }
  long long val = me(2 * m - 1, n), temp = me(m, n);
  val = (val - temp + mod) % mod;
  val = (val * m) % mod;
  val = (val * mi(m - 1));
  val = (val + temp) % mod;
  printf("%lld", val);
  return 0;
}

### Prompt

Please formulate 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 = 1e9 + 7;
const int N = 1e6 + 4;
long long me(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return res;
}
long long mi(long long n) { return me(n, mod - 2); }
int main() {
  long long n, m;
  scanf("%lld", &n);
  scanf("%lld", &m);
  if (m == 1) {
    printf("%lld", n + 1);
    return 0;
  }
  long long val = me(2 * m - 1, n), temp = me(m, n);
  val = (val - temp + mod) % mod;
  val = (val * m) % mod;
  val = (val * mi(m - 1));
  val = (val + temp) % mod;
  printf("%lld", val);
  return 0;
}

```