#include <bits/stdc++.h>
using namespace std;
const int MAX = 1123456;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long potencia(long long m, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    return potencia(m * m % MOD, n / 2);
  }
  return (potencia(m * m % MOD, n / 2) * m) % MOD;
}
int main() {
  long long n, m;
  scanf("%I64d %I64d", &n, &m);
  long long ans = 0;
  if (m == 1) {
    printf("%I64d", n + 1);
    return 0;
  }
  ans += potencia(m, n) % MOD;
  long long aux = (m * potencia(m - 1, n - 1)) % MOD;
  ans = (ans + aux) % MOD;
  for (long long s = 1; s < n; s++) {
    aux *= m;
    aux %= MOD;
    aux *= n - s + 1;
    aux %= MOD;
    aux *= potencia(s, MOD - 2);
    aux %= MOD;
    aux *= potencia(m - 1, MOD - 2);
    aux %= MOD;
    ans = (ans + aux) % MOD;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 int MAX = 1123456;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long potencia(long long m, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    return potencia(m * m % MOD, n / 2);
  }
  return (potencia(m * m % MOD, n / 2) * m) % MOD;
}
int main() {
  long long n, m;
  scanf("%I64d %I64d", &n, &m);
  long long ans = 0;
  if (m == 1) {
    printf("%I64d", n + 1);
    return 0;
  }
  ans += potencia(m, n) % MOD;
  long long aux = (m * potencia(m - 1, n - 1)) % MOD;
  ans = (ans + aux) % MOD;
  for (long long s = 1; s < n; s++) {
    aux *= m;
    aux %= MOD;
    aux *= n - s + 1;
    aux %= MOD;
    aux *= potencia(s, MOD - 2);
    aux %= MOD;
    aux *= potencia(m - 1, MOD - 2);
    aux %= MOD;
    ans = (ans + aux) % MOD;
  }
  printf("%I64d\n", ans);
  return 0;
}

```