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 <algorithm>
#include <vector>
#include <queue>
 
using namespace std;
 
#define DEBUG(x) // cerr << #x << " = " << x << endl
 
const int INF = (int)1e9;
 
int main() {
	int N, M, S, T; cin >> N >> M >> S >> T;
	S--, T--;
	vector<vector<int> > E(N);
	for(int i = 0; i < M; i++) {
		int a, b; cin >> a >> b;
		a--, b--;
		E[a].push_back(b);
		E[b].push_back(a);
	}
 
	queue<int> Q;
 
	vector<int> distS(N, INF);
	vector<int> distSumS(N, 0);
 
	Q.push(S);
	distS[S] = 0;
	distSumS[0]++;
	while(!Q.empty()) {
		int cur = Q.front(); Q.pop();
		for(int i = 0; i < (int)E[cur].size(); i++) {
			int to = E[cur][i];
			if(distS[cur] + 1 < distS[to]) {
				Q.push(to);
				distS[to] = distS[cur] + 1;
				distSumS[distS[to]]++;
			}
		}
	}
 
	for(int i = 0; i < N; i++) {
		// cerr << "distS[" << i << "]=" << distS[i] << endl;
		// cerr << "distSumS[" << i << "]=" << distSumS[i] << endl;
	}
 
	Q = queue<int>();
	vector<int> distT(N, INF);
	vector<int> distSumT(N, 0);
 
	Q.push(T);
	distT[T] = 0;
	distSumT[0]++;
	while(!Q.empty()) {
		int cur = Q.front(); Q.pop();
		for(int i = 0; i < (int)E[cur].size(); i++) {
			int to = E[cur][i];
			if(distT[cur] + 1 < distT[to]) {
				Q.push(to);
				distT[to] = distT[cur] + 1;
				distSumT[distT[to]]++;
			}
		}
	}
 
	for(int i = 0; i < N; i++) {
		// cerr << "distT[" << i << "]=" << distT[i] << endl;
		// cerr << "distSumT[" << i << "]=" << distSumT[i] << endl;
	}
 
	int minDist = distS[T];
	DEBUG(minDist);
	long long ans = 0;
	for(int i = 0; i <= minDist - 2; i++) {
		ans += (long long)distSumS[i] * (long long)distSumT[minDist - i - 2];
	}
	cout << ans << endl;
}