#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
void ADD(long long& x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
long long fac[SIZE], inv[SIZE];
long long mypow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    y >>= 1;
    x = x * x % MOD;
  }
  return res;
}
long long C(int x, int y) {
  if (y < 0 || y > x) return 0;
  return fac[x] * inv[y] % MOD * inv[x - y] % MOD;
}
void build() {
  assert(MOD >= SIZE);
  fac[0] = 1;
  for (int i = (1); i < (SIZE); ++i) fac[i] = fac[i - 1] * i % MOD;
  inv[SIZE - 1] = mypow(fac[SIZE - 1], MOD - 2);
  for (int i = SIZE - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
int main() {
  build();
  int n, m;
  scanf("%d%d", &n, &m);
  long long now = 0, an = 0, tmp = 1;
  for (int i = n; i >= 0; i--) {
    ADD(now, C(n, i) * tmp);
    tmp = tmp * (m - 1) % MOD;
    ADD(an, now * mypow(m, i));
  }
  cout << an;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
void ADD(long long& x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
long long fac[SIZE], inv[SIZE];
long long mypow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    y >>= 1;
    x = x * x % MOD;
  }
  return res;
}
long long C(int x, int y) {
  if (y < 0 || y > x) return 0;
  return fac[x] * inv[y] % MOD * inv[x - y] % MOD;
}
void build() {
  assert(MOD >= SIZE);
  fac[0] = 1;
  for (int i = (1); i < (SIZE); ++i) fac[i] = fac[i - 1] * i % MOD;
  inv[SIZE - 1] = mypow(fac[SIZE - 1], MOD - 2);
  for (int i = SIZE - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
int main() {
  build();
  int n, m;
  scanf("%d%d", &n, &m);
  long long now = 0, an = 0, tmp = 1;
  for (int i = n; i >= 0; i--) {
    ADD(now, C(n, i) * tmp);
    tmp = tmp * (m - 1) % MOD;
    ADD(an, now * mypow(m, i));
  }
  cout << an;
  return 0;
}

```