Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2
#include <bits/stdc++.h>
using namespace std;

int n, m, s, t;
vector<int> g[100000];
int d1[100000], d2[100000];
int c1[100010], c2[100010];

void bfs(int dist[], int s)
{
	queue<int> q;
	fill_n(dist, 100000, -1);
	dist[s] = 0;
	q.push(s);
	while (q.size()){
		int v = q.front(); q.pop();
		for (int to : g[v]){
			if (dist[to] >= 0) continue;
			dist[to] = dist[v] + 1;
			q.push(to);
		}
	}
}

int main()
{
	cin >> n >> m >> s >> t, s--, t--;
	for (int i = 0; i < m; i++){
		int x, y;
		cin >> x >> y, x--, y--;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	
	bfs(d1, s);
	bfs(d2, t);

	for (int i = 0; i < n; i++){
		if (d1[i] >= 0) c1[d1[i]]++;
		if (d2[i] >= 0) c2[d2[i]]++;
	}

	long long res = 0;
	for (int i = 0; i < d1[t] - 1; i++){
		res += (long long)c1[i] * c2[d1[t] - 2 - i];
	}

	cout << res << endl;
}