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;
using ll = long long;
using graph = vector<vector<int>>;

const int INF = 1e9;

vector<int> bfs(int s, const graph& G) {
	vector<int> d(G.size(), INF);
	queue<int> q;
	q.push(s);
	d[s] = 0;
	while (!q.empty()) {
		int v = q.front(); q.pop();
		for (auto to : G[v]) if (d[to] == INF) {
			q.push(to);
			d[to] = d[v] + 1;
		}
	}
	return d;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	int N, M, s, t;
	cin >> N >> M >> s >> t; --s, --t;
	graph G(N);
	for (int i = 0; i < M; i++) {
		int u, v;
		cin >> u >> v; --u, --v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	auto d1 = bfs(s, G);
	auto d2 = bfs(t, G);
	int dist = d1[t];
	sort(d1.begin(), d1.end());
	sort(d2.begin(), d2.end());
	ll res = 0;
	for (int i = 0; i < N; i++) {
		if (d1[i] <= dist - 2) {
			res += upper_bound(d2.begin(), d2.end(), dist - 2 - d1[i])
				 - lower_bound(d2.begin(), d2.end(), dist - 2 - d1[i]);
		}
	}
	cout << res << endl;
	return 0;
}

