Problem: Once, during a lesson, Sasha got bored and decided to talk with his friends. Suddenly, he saw Kefa. Since we can talk endlessly about Kefa, we won't even start doing that. The conversation turned to graphs. Kefa promised Sasha to tell him about one interesting fact from graph theory if Sasha helps Kefa to count the number of beautiful trees. 

In this task, a tree is a weighted connected graph, consisting of n vertices and n-1 edges, and weights of edges are integers from 1 to m. Kefa determines the beauty of a tree as follows: he finds in the tree his two favorite vertices — vertices with numbers a and b, and counts the distance between them. The distance between two vertices x and y is the sum of weights of edges on the simple path from x to y. If the distance between two vertices a and b is equal to m, then the tree is beautiful.

Sasha likes graph theory, and even more, Sasha likes interesting facts, that's why he agreed to help Kefa. Luckily, Sasha is familiar with you the best programmer in Byteland. Help Sasha to count the number of beautiful trees for Kefa. Two trees are considered to be distinct if there is an edge that occurs in one of them and doesn't occur in the other one. Edge's weight matters.

Kefa warned Sasha, that there can be too many beautiful trees, so it will be enough to count the number modulo 10^9 + 7.

Input

The first line contains four integers n, m, a, b (2 ≤ n ≤ 10^6, 1 ≤ m ≤ 10^6, 1 ≤ a, b ≤ n, a ≠ b) — the number of vertices in the tree, the maximum weight of an edge and two Kefa's favorite vertices.

Output

Print one integer — the number of beautiful trees modulo 10^9+7.

Examples

Input


3 2 1 3


Output


5


Input


3 1 1 2


Output


2


Input


5 15 1 5


Output


345444

Note

There are 5 beautiful trees in the first example:

<image>

In the second example the following trees are beautiful:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = (1000000007LL);
inline long long Mod(long long a, long long b) { return (a % b); }
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b, swap(a, b);
  }
  return a;
}
void read(vector<long long> &w, long long n) {
  w.resize(n);
  for (long long i = 0; i < n; i++) cin >> w[i];
}
void print(vector<long long> &w) {
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (i == (long long)(w).size() - 1)
      cout << w[i] << "\n";
    else
      cout << w[i] << " ";
  }
}
long long prodmod(vector<long long> w);
void crivo(vector<long long> &w, long long TAM);
void inverso(vector<long long> &inv, long long TAM);
void fat_inverso(vector<long long> &inv, long long TAM);
long long choose(long long a, long long b);
void get_fat();
long long fat[2000000];
long long n, m, a, b;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> a >> b;
  long long ans = 0;
  fat[0] = 1;
  for (long long i = 1; i < 2000000; i++) fat[i] = (i * fat[i - 1]) % mod;
  for (long long k = 2; k <= n; k++) {
    long long A = (choose(n - 2, k - 2) * fat[k - 2]) % mod;
    long long B = (choose(m - 1, k - 2) * poww(m, n - k)) % mod;
    long long C =
        (n - k - 1 >= 0 ? (k * poww(n, n - k - 1)) : (k * poww(n, mod - 2))) %
        mod;
    A = prodmod({A, B, C});
    ans += A;
    ans %= mod;
  }
  cout << ans << "\n";
}
long long prodmod(vector<long long> w) {
  long long curr = 1;
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (w[i] >= mod) w[i] %= mod;
    curr = (curr * w[i]) % mod;
    if (curr < 0) curr += mod;
  }
  return curr;
}
void get_fat() {
  if (fat[0] != 0) return;
  fat[0] = 1;
  for (long long i = 1; i < 2000000; i++) fat[i] = (i * fat[i - 1]) % mod;
}
long long choose(long long a, long long b) {
  get_fat();
  if (b > a) return 0;
  if (!b) return 1;
  long long A = fat[a];
  long long B = (poww(fat[b], mod - 2) * poww(fat[a - b], mod - 2)) % mod;
  A = (A * B) % mod;
  return A;
}
void inverso(vector<long long> &inv, long long TAM) {
  inv.resize(TAM);
  inv[1] = 1;
  for (long long i = 2; i < TAM; ++i) {
    inv[i] = (mod - (mod / i) * inv[mod % i] % mod) % mod;
  }
}
void fat_inverso(vector<long long> &inv, long long TAM) {
  inverso(inv, TAM);
  for (long long i = 2; i < TAM; i++) inv[i] = (inv[i] * inv[i - 1]) % mod;
}
void crivo(vector<long long> &w, long long TAM) {
  w.resize(TAM + 1);
  w[0] = 1;
  for (long long i = 2; i < TAM; i++)
    if (!w[i])
      for (long long j = 2 * i; j < TAM; j += i) w[j] = 1;
}