#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << endl;
  err(++it, args...);
}
const long long MOD = 1e9 + 7;
long long modExp(long long b, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) {
      res = (res * b) % MOD;
    }
    b = (b * b) % MOD;
    p >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    ans = ((2 * m - 1) * ans + modExp(m, i - 1)) % MOD;
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << endl;
  err(++it, args...);
}
const long long MOD = 1e9 + 7;
long long modExp(long long b, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) {
      res = (res * b) % MOD;
    }
    b = (b * b) % MOD;
    p >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    ans = ((2 * m - 1) * ans + modExp(m, i - 1)) % MOD;
  }
  cout << ans;
  return 0;
}

```