#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5, MOD = 1e9 + 7;
long long ipow(long long x, long long p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  return ipow(x * x % MOD, p / 2) * ipow(x, p % 2) % MOD;
}
long long inverse(long long x) { return ipow(x, MOD - 2); }
long long fact[N], inv_fact[N];
long long comb(long long n, long long m) {
  return fact[n] * inv_fact[m] % MOD * inv_fact[n - m] % MOD;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = (fact[i - 1] * i) % MOD;
  inv_fact[N - 1] = inverse(fact[N - 1]);
  for (int i = N - 2; i >= 0; --i)
    inv_fact[i] = (inv_fact[i + 1] * (i + 1)) % MOD;
  long long n, m;
  cin >> n >> m;
  long long ans = ipow(m, n), all = ipow(m, n);
  for (long long k = 1; k <= n; ++k) {
    all = (all - ipow(m - 1, n - (k - 1)) * comb(n, (k - 1))) % MOD;
    ans = (ans + ipow(m, k) * all) % MOD;
  }
  cout << (ans % MOD + MOD) % MOD << '\n';
}

### 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 N = 2e6 + 5, MOD = 1e9 + 7;
long long ipow(long long x, long long p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  return ipow(x * x % MOD, p / 2) * ipow(x, p % 2) % MOD;
}
long long inverse(long long x) { return ipow(x, MOD - 2); }
long long fact[N], inv_fact[N];
long long comb(long long n, long long m) {
  return fact[n] * inv_fact[m] % MOD * inv_fact[n - m] % MOD;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = (fact[i - 1] * i) % MOD;
  inv_fact[N - 1] = inverse(fact[N - 1]);
  for (int i = N - 2; i >= 0; --i)
    inv_fact[i] = (inv_fact[i + 1] * (i + 1)) % MOD;
  long long n, m;
  cin >> n >> m;
  long long ans = ipow(m, n), all = ipow(m, n);
  for (long long k = 1; k <= n; ++k) {
    all = (all - ipow(m - 1, n - (k - 1)) * comb(n, (k - 1))) % MOD;
    ans = (ans + ipow(m, k) * all) % MOD;
  }
  cout << (ans % MOD + MOD) % MOD << '\n';
}

```