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 <iostream>
#include <string>
#include <list>
#include <queue>
#define llint long long
#define inf 1e18

using namespace std;

llint n, m, s, t;
vector<llint> G[100005];
llint distS[100005], distT[100005];
llint cntS[100005], cntT[100005];

void bfs(llint S, llint dist[])
{
	for(int i = 1; i <= n; i++) dist[i] = inf;
	dist[S] = 0;
	
	queue<llint> Q;
	Q.push(S);
	
	llint v;
	while(Q.size()){
		v = Q.front();
		Q.pop();
		for(int i = 0 ;i < G[v].size(); i++){
			llint u = G[v][i];
			if(dist[u] < inf) continue;
			Q.push(u);
			dist[u] = dist[v] + 1;
		}
	}
}

int main(void)
{
	cin >> n >> m >> s >> t;
	llint u, v;
	for(int i = 1; i <= m; i++){
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	bfs(s, distS);
	bfs(t, distT);
	llint d = distS[t];
	
	for(int i = 1; i <= n; i++){
		if(distS[i] < inf/2) cntS[distS[i]]++;
		if(distT[i] < inf/2) cntT[distT[i]]++;
	}
	
	llint ans = 0;
	for(int i = 0; i <= d-2; i++){
		ans += cntS[i]*cntT[d-2-i];
	}
	cout << ans << endl;
	
	return 0;
}
