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<queue>
#include<vector>
#include<functional>

using namespace std;
typedef long long ll;
typedef pair<ll, int> P;

int n, m, s, t;

ll dis_from_s[100000], dis_from_t[100000];
ll cnt_dis_s[200000], cnt_dis_t[200000];
vector<vector<int>> edge;

void djk(int pos, ll dis[], ll cnt[]) {
	
	priority_queue<P, vector<P>, greater<P>> pq;
	pq.push(P(0, pos));
	dis[pos] = 0;
	while (pq.size()) {
		auto tmp = pq.top(); pq.pop();
		if (dis[tmp.second] < tmp.first)continue;

		cnt[tmp.first]++;

		for (int i = 0; i < edge[tmp.second].size(); i++) {
			int next = edge[tmp.second][i];
			if (dis[next] > dis[tmp.second] + 1) {
				dis[next] = dis[tmp.second] + 1;
				pq.push(P(dis[tmp.second] + 1, next));
			}
		}
	}
}

int main() {

	cin >> n >> m >> s >> t;
	s--, t--;
	edge.resize(n);
	for (int i = 0; i < n; i++)dis_from_s[i] = 100000000, dis_from_t[i] = 1000000000;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		edge[a].push_back(b);
		edge[b].push_back(a);
	}

	djk(s, dis_from_s, cnt_dis_s), djk(t, dis_from_t, cnt_dis_t);
	
	ll ans = 0;
	
	int s_to_t = dis_from_s[t];

	for (int i = 0; i <= s_to_t - 2; i++) {
		ans += (cnt_dis_s[i] * cnt_dis_t[s_to_t - 2 - i]);
	}
	cout << ans << endl;
	return 0;
}
