#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1010101;
int n, m;
int pw[2][N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n >> m;
  pw[0][0] = pw[1][0] = 1;
  for (int i = (1); i < (n + 1); i++) {
    pw[0][i] = mul(pw[0][i - 1], m);
    pw[1][i] = mul(pw[1][i - 1], m * 2 - 1);
  }
  int ans = pw[0][n];
  for (int i = (1); i < (n + 1); i++)
    ans = add(ans, mul(pw[0][n - i + 1], pw[1][i - 1]));
  cout << ans << "\n";
  return 0;
}

### Prompt

Your task is to create 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 P = 1e9 + 7;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1010101;
int n, m;
int pw[2][N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n >> m;
  pw[0][0] = pw[1][0] = 1;
  for (int i = (1); i < (n + 1); i++) {
    pw[0][i] = mul(pw[0][i - 1], m);
    pw[1][i] = mul(pw[1][i - 1], m * 2 - 1);
  }
  int ans = pw[0][n];
  for (int i = (1); i < (n + 1); i++)
    ans = add(ans, mul(pw[0][n - i + 1], pw[1][i - 1]));
  cout << ans << "\n";
  return 0;
}

```