#include <bits/stdc++.h>
const int MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int n, m;
int f[MAXN];
inline void add(int &x, int y) {
  x += y - ha;
  x += x >> 31 & ha;
}
struct Node {
  int a[2][2];
  Node() { memset(a, 0, sizeof(a)); }
  inline Node operator*(const Node &t) const {
    Node res;
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j) {
        long long x = 0;
        for (int k = 0; k <= 1; ++k) x += (long long)a[i][k] * t.a[k][j];
        res.a[i][j] = x % ha;
      }
    return res;
  }
};
inline Node qpow(Node a, int n) {
  Node res;
  res.a[0][0] = res.a[1][1] = 1;
  while (n) {
    if (n & 1) res = res * a;
    a = a * a;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  Node A;
  A.a[1][0] = 1;
  A.a[0][0] = 3ll * m - 1;
  A.a[0][1] = (ha - 2ll * m * m % ha + m) % ha;
  A = qpow(A, n - 1);
  int ans = 1ll * A.a[0][0] * (3 * m - 1) % ha;
  add(ans, A.a[0][1]);
  (ans += ha - qpow(m, n)) %= ha;
  ans = 1ll * ans * m % ha * qpow(2 * m - 1) % ha;
  (ans += qpow(m, n)) %= ha;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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>
const int MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int n, m;
int f[MAXN];
inline void add(int &x, int y) {
  x += y - ha;
  x += x >> 31 & ha;
}
struct Node {
  int a[2][2];
  Node() { memset(a, 0, sizeof(a)); }
  inline Node operator*(const Node &t) const {
    Node res;
    for (int i = 0; i <= 1; ++i)
      for (int j = 0; j <= 1; ++j) {
        long long x = 0;
        for (int k = 0; k <= 1; ++k) x += (long long)a[i][k] * t.a[k][j];
        res.a[i][j] = x % ha;
      }
    return res;
  }
};
inline Node qpow(Node a, int n) {
  Node res;
  res.a[0][0] = res.a[1][1] = 1;
  while (n) {
    if (n & 1) res = res * a;
    a = a * a;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  Node A;
  A.a[1][0] = 1;
  A.a[0][0] = 3ll * m - 1;
  A.a[0][1] = (ha - 2ll * m * m % ha + m) % ha;
  A = qpow(A, n - 1);
  int ans = 1ll * A.a[0][0] * (3 * m - 1) % ha;
  add(ans, A.a[0][1]);
  (ans += ha - qpow(m, n)) %= ha;
  ans = 1ll * ans * m % ha * qpow(2 * m - 1) % ha;
  (ans += qpow(m, n)) %= ha;
  printf("%d\n", ans);
  return 0;
}

```