#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1100100;
const int MODP = 1e9 + 7;
long long log_pow(long long b, long long e) {
  long long r = 1;
  while (e) {
    if (e & 1) r = r * b % MODP;
    b = b * b % MODP;
    e >>= 1;
  }
  return r;
}
long long f[MAXN], invf[MAXN];
long long comb(long long n, long long k) {
  if (k > n) return 0;
  return f[n] * invf[k] % MODP * invf[n - k] % MODP;
}
long long n, m, sol;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  f[0] = 1;
  for (int i = 1; i < MAXN; i++) f[i] = f[i - 1] * i % MODP;
  invf[MAXN - 1] = log_pow(f[MAXN - 1], MODP - 2);
  for (int i = MAXN - 2; i >= 0; i--) invf[i] = invf[i + 1] * (i + 1) % MODP;
  cin >> n >> m;
  sol = log_pow(m, n);
  for (int s = 1; s <= n; s++) {
    long long x =
        log_pow(m, s) * log_pow(m - 1, n - s) % MODP * comb(n, s - 1) % MODP;
    sol = (sol + x) % MODP;
  }
  cout << sol << '\n';
  ;
}

### Prompt

Generate 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 MAXN = 1100100;
const int MODP = 1e9 + 7;
long long log_pow(long long b, long long e) {
  long long r = 1;
  while (e) {
    if (e & 1) r = r * b % MODP;
    b = b * b % MODP;
    e >>= 1;
  }
  return r;
}
long long f[MAXN], invf[MAXN];
long long comb(long long n, long long k) {
  if (k > n) return 0;
  return f[n] * invf[k] % MODP * invf[n - k] % MODP;
}
long long n, m, sol;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  f[0] = 1;
  for (int i = 1; i < MAXN; i++) f[i] = f[i - 1] * i % MODP;
  invf[MAXN - 1] = log_pow(f[MAXN - 1], MODP - 2);
  for (int i = MAXN - 2; i >= 0; i--) invf[i] = invf[i + 1] * (i + 1) % MODP;
  cin >> n >> m;
  sol = log_pow(m, n);
  for (int s = 1; s <= n; s++) {
    long long x =
        log_pow(m, s) * log_pow(m - 1, n - s) % MODP * comb(n, s - 1) % MODP;
    sol = (sol + x) % MODP;
  }
  cout << sol << '\n';
  ;
}

```