#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e17 + 7;
const long long mod = 1e9 + 7;
long long binpow(long long a, long long b) {
  long long res = 1;
  for (; b; b /= 2, a = a * a % mod)
    if (b & 1) res = res * a % mod;
  return res;
}
long long reverse(long long a) { return binpow(a, mod - 2); }
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  long long q = n * binpow(m, n - 1) % mod;
  if (m != 2 * m - 1)
    q = (binpow(2 * m - 1, n) - binpow(m, n)) * reverse(m - 1) % mod;
  long long res = m * q % mod + binpow(m, n) % mod;
  cout << (res + mod) % mod;
}

### 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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e17 + 7;
const long long mod = 1e9 + 7;
long long binpow(long long a, long long b) {
  long long res = 1;
  for (; b; b /= 2, a = a * a % mod)
    if (b & 1) res = res * a % mod;
  return res;
}
long long reverse(long long a) { return binpow(a, mod - 2); }
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  long long q = n * binpow(m, n - 1) % mod;
  if (m != 2 * m - 1)
    q = (binpow(2 * m - 1, n) - binpow(m, n)) * reverse(m - 1) % mod;
  long long res = m * q % mod + binpow(m, n) % mod;
  cout << (res + mod) % mod;
}

```