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;

#define for_(i,a,b) for(int i=(a);i<(b);++i)
#define allof(a) (a).begin(),(a).end()

typedef long long lint;

struct State { int v, step; };

int N, M, term[2], dst[2][100010];
vector< vector< int > > adj;
vector< int > sdl[2][100010];

int bfs() {
	for_(i,0,2) {
		queue< State > que;
		que.push(State{term[i], 0});
		
		vector< bool > vis(N + 1, false);
		vis[term[i]] = true;
		
		while (!que.empty()) {
			State s = que.front(); que.pop();
			dst[i][s.v] = s.step;
			sdl[i][s.step].push_back(s.v);
			
			for (int v : adj[s.v]) {
				if (vis[v]) continue;
				vis[v] = true;
				que.push(State{v, s.step + 1});
			}
		}
	}
	
	return dst[0][term[1]];
}

void solve() {
	int min_dst = bfs();
	
	lint ans = 0;
	
	for_(step,0,N) ans += (lint)sdl[0][step].size() * (lint)sdl[1][min_dst - step - 2].size();
	
	cout << ans << endl;
}

int main() {
	cin >> N >> M >> term[0] >> term[1];
	
	adj.assign(N+1, vector< int >());
	
	for_(i,0,M) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	
	solve();
}