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;

typedef vector<vector<int>> graph;

vector<int> bfs(int s, const graph &G) {
	constexpr int INF = (1 << 30);
	const int n = G.size();

	vector<int> dist(n, INF);
	queue<int> que;
	dist[s] = 0;
	que.push(s);

	while(!que.empty()) {
		const int v = que.front();
		que.pop();

		for(const auto &to : G[v]) {
			if(dist[to] > dist[v] + 1) {
				dist[to] = dist[v] + 1;
				que.push(to);
			}
		}
	}
	return dist;
}

int main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int n, m, s, t;
	cin >> n >> m >> s >> t;
	--s; --t;

	graph G(n);
	for(int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		G[a].emplace_back(b);
		G[b].emplace_back(a);
	}

	const auto dist_from_s = bfs(s, G);
	const auto dist_from_t = bfs(t, G);
	const auto dist_st = dist_from_s[t];

	unordered_map<int, int> cnt;
	for(const auto &d : dist_from_t) {
		++cnt[d];
	}

	long long ans = 0;
	for(const auto &e : dist_from_s) {
		ans += cnt[dist_st - e - 2];
	}
	cout << ans << endl;

	return EXIT_SUCCESS;
}