#include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
const long long mod = 1e9 + 7;
long long extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0 && b == 0) return -1;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extend_gcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
long long mod_reverse(long long a, long long n) {
  long long x, y;
  long long d = extend_gcd(a, n, x, y);
  if (d == 1)
    return (x % n + n) % n;
  else
    return -1;
}
long long pow(long long a, int k) {
  long long ret = 1, tmp = a;
  while (k) {
    if (k & 1) ret = (ret * tmp) % mod;
    tmp = (tmp * tmp) % mod;
    k >>= 1;
  }
  return ret;
}
int main() {
  int n, m;
  cin >> n >> m;
  long long a = m - 1;
  a = (a * mod_reverse(m, mod)) % mod;
  a = (a + 1) % mod;
  long long k = pow((long long)m, n);
  long long t = 0;
  for (int i = 0; i < n; i++) {
    t = (t + pow(a, i)) % mod;
  }
  t = (t * k + k) % mod;
  cout << t << endl;
  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;
const int maxn = 300005;
const long long mod = 1e9 + 7;
long long extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0 && b == 0) return -1;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extend_gcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
long long mod_reverse(long long a, long long n) {
  long long x, y;
  long long d = extend_gcd(a, n, x, y);
  if (d == 1)
    return (x % n + n) % n;
  else
    return -1;
}
long long pow(long long a, int k) {
  long long ret = 1, tmp = a;
  while (k) {
    if (k & 1) ret = (ret * tmp) % mod;
    tmp = (tmp * tmp) % mod;
    k >>= 1;
  }
  return ret;
}
int main() {
  int n, m;
  cin >> n >> m;
  long long a = m - 1;
  a = (a * mod_reverse(m, mod)) % mod;
  a = (a + 1) % mod;
  long long k = pow((long long)m, n);
  long long t = 0;
  for (int i = 0; i < n; i++) {
    t = (t + pow(a, i)) % mod;
  }
  t = (t * k + k) % mod;
  cout << t << endl;
  return 0;
}

```