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++)
#define INF 10000000
using namespace std;
typedef pair<int,int> P;

int n, m, s, t;
int dist[2][100000];
vector<int> e[100000];
long long num[2][100001];

int main(){
	cin >> n >> m >> s >> t;
	rep(u,2) rep(i,100000) dist[u][i] = INF;
	s--; t--;
	rep(i,m){
		int x, y;
		cin >> x >> y;
		x--; y--;
		e[x].push_back(y);
		e[y].push_back(x);
	}
	queue<P> que;
	dist[0][s] = 0;
	num[0][0]++;
	que.push(P(0,s));
	while(true){
		if(que.size() == 0) break;
		P q = que.front();
		que.pop();
		rep(i,e[q.second].size()){
			if(dist[0][e[q.second][i]] != INF) continue;
			dist[0][e[q.second][i]] = q.first+1;
			num[0][q.first+1]++;
			que.push(P(q.first+1,e[q.second][i]));
		}
	}
	dist[1][t] = 0;
	num[1][0]++;
	que.push(P(0,t));
	while(true){
		if(que.size() == 0) break;
		P q = que.front();
		que.pop();
		rep(i,e[q.second].size()){
			if(dist[1][e[q.second][i]] != INF) continue;
			dist[1][e[q.second][i]] = q.first+1;
			num[1][q.first+1]++;
			que.push(P(q.first+1,e[q.second][i]));
		}
	}
	long long ans = 0;
	int D = dist[0][t];
	for(int i = 0; i <= D-2; i++){
		ans += num[0][i]*num[1][D-2-i];
	}
	cout << ans << endl;
}