#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long mult_mod(long long a, long long b, long long c) {
  a %= c;
  b %= c;
  long long ret = 0;
  long long tmp = a;
  while (b) {
    if (b & 1) {
      ret += tmp;
      if (ret > c) ret -= c;
    }
    tmp <<= 1;
    if (tmp > c) tmp -= c;
    b >>= 1;
  }
  return ret;
}
long long pow_mod(long long a, long long n, long long mod) {
  long long ret = 1;
  long long temp = a % mod;
  while (n) {
    if (n & 1) ret = mult_mod(ret, temp, mod);
    temp = mult_mod(temp, temp, mod);
    n >>= 1;
  }
  return ret;
}
int main() {
  int n, m;
  cin >> n;
  cin >> m;
  if (m == 1) {
    cout << (n + 1) << endl;
    return 0;
  }
  long long ans = 0;
  ans = pow_mod(m, n, mod);
  long long ans1 =
      (pow_mod(2 * m - 1, n, mod) - pow_mod(m, n, mod) + mod) % mod;
  ans1 = mult_mod(ans1, m, mod);
  long long ans2 = pow_mod(m - 1, mod - 2, mod);
  ans1 = mult_mod(ans1, ans2, mod);
  ans = (ans1 + ans) % mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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 = 1000000007;
long long mult_mod(long long a, long long b, long long c) {
  a %= c;
  b %= c;
  long long ret = 0;
  long long tmp = a;
  while (b) {
    if (b & 1) {
      ret += tmp;
      if (ret > c) ret -= c;
    }
    tmp <<= 1;
    if (tmp > c) tmp -= c;
    b >>= 1;
  }
  return ret;
}
long long pow_mod(long long a, long long n, long long mod) {
  long long ret = 1;
  long long temp = a % mod;
  while (n) {
    if (n & 1) ret = mult_mod(ret, temp, mod);
    temp = mult_mod(temp, temp, mod);
    n >>= 1;
  }
  return ret;
}
int main() {
  int n, m;
  cin >> n;
  cin >> m;
  if (m == 1) {
    cout << (n + 1) << endl;
    return 0;
  }
  long long ans = 0;
  ans = pow_mod(m, n, mod);
  long long ans1 =
      (pow_mod(2 * m - 1, n, mod) - pow_mod(m, n, mod) + mod) % mod;
  ans1 = mult_mod(ans1, m, mod);
  long long ans2 = pow_mod(m - 1, mod - 2, mod);
  ans1 = mult_mod(ans1, ans2, mod);
  ans = (ans1 + ans) % mod;
  cout << ans << endl;
  return 0;
}

```