#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
long long int pw(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int tmp = pw(x, n / 2);
  if (n % 2)
    return (tmp * tmp % mod) * x % mod;
  else
    return tmp * tmp % mod;
}
long long int mod_inv(long long int x) { return pw(x, mod - 2); }
long long int N, M;
int main() {
  scanf("%I64d %I64d", &N, &M);
  long long int tmp = 1, k = (1 + (M - 1) * mod_inv(M) % mod) % mod;
  long long int ans = 0;
  for (int i = 0; i < N; i++) {
    ans += tmp;
    ans %= mod;
    tmp *= k;
    tmp %= mod;
  }
  ans += 1;
  ans %= mod;
  ans *= pw(M, N);
  ans %= mod;
  printf("%I64d", ans);
}

### 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 int mod = 1000000007;
long long int pw(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int tmp = pw(x, n / 2);
  if (n % 2)
    return (tmp * tmp % mod) * x % mod;
  else
    return tmp * tmp % mod;
}
long long int mod_inv(long long int x) { return pw(x, mod - 2); }
long long int N, M;
int main() {
  scanf("%I64d %I64d", &N, &M);
  long long int tmp = 1, k = (1 + (M - 1) * mod_inv(M) % mod) % mod;
  long long int ans = 0;
  for (int i = 0; i < N; i++) {
    ans += tmp;
    ans %= mod;
    tmp *= k;
    tmp %= mod;
  }
  ans += 1;
  ans %= mod;
  ans *= pw(M, N);
  ans %= mod;
  printf("%I64d", ans);
}

```