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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int d[100000];
int c1[100000], c2[100000];
vector<int>E[100000];
void bfs(int*c, int s) {
	memset(d, -1, sizeof(d));
	queue<int>que;
	d[s] = 0; que.push(s);
	while (!que.empty()) {
		int p = que.front(); que.pop();
		c[d[p]]++;
		for (int v : E[p]) {
			if (d[v] == -1) {
				d[v] = d[p] + 1; que.push(v);
			}
		}
	}
}
int main() {
	int n, m, s, t; scanf("%d%d%d%d", &n, &m, &s, &t); s--; t--;
	rep(i, m) {
		int x, y; scanf("%d%d", &x, &y); x--; y--;
		E[x].push_back(y); E[y].push_back(x);
	}
	bfs(c1, t); bfs(c2, s);
	ll ans = 0;
	for (int i = 0; i <= d[t] - 2; i++) {
		ans += (ll)c1[i] * c2[d[t] - 2 - i];
	}
	printf("%lld\n", ans);
}