#include <bits/stdc++.h>
using namespace std;
long long n, m, a[1000050], b[1000050], c[1000050], ans;
long long Pow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return Pow(x, 1000000007 - 2); }
int main() {
  scanf("%I64d%I64d", &n, &m);
  a[0] = b[0] = c[0] = 1;
  for (register int i = 1; i <= n; ++i)
    a[i] = a[i - 1] * m % 1000000007, b[i] = b[i - 1] * (m - 1) % 1000000007,
    c[i] = c[i - 1] * (n - i + 1) % 1000000007 * inv(i) % 1000000007;
  for (register int i = 1; i <= n; ++i)
    ans += a[i] * b[n - i] % 1000000007 * c[i - 1] % 1000000007,
        ans %= 1000000007;
  ans += a[n];
  printf("%I64d", ans % 1000000007);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long n, m, a[1000050], b[1000050], c[1000050], ans;
long long Pow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return Pow(x, 1000000007 - 2); }
int main() {
  scanf("%I64d%I64d", &n, &m);
  a[0] = b[0] = c[0] = 1;
  for (register int i = 1; i <= n; ++i)
    a[i] = a[i - 1] * m % 1000000007, b[i] = b[i - 1] * (m - 1) % 1000000007,
    c[i] = c[i - 1] * (n - i + 1) % 1000000007 * inv(i) % 1000000007;
  for (register int i = 1; i <= n; ++i)
    ans += a[i] * b[n - i] % 1000000007 * c[i - 1] % 1000000007,
        ans %= 1000000007;
  ans += a[n];
  printf("%I64d", ans % 1000000007);
  return 0;
}

```