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
#define  _CRT_SECURE_NO_WARNINGS
#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-11L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL

#define MOD 998244353LL
#define seg_size 262144 * 4LL
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

unsigned long xor128() {
	static unsigned long x = 123456789, y = 362436069, z = 521288629, w = time(NULL);
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
void init() {
	iostream::sync_with_stdio(false);
	cout << fixed << setprecision(20);
}

#define int ll

int dp[2][200000];
vector<int> vertexs[200000];
int cnt[2][200000];
vector<int> check[2][200000];
void solve(){
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	REP(i, n) {
		dp[0][i+1] = 1e9;
		dp[1][i+1] = 1e9;
	}
	dp[0][s] = 0;
	dp[1][t] = 0;
	REP(i, m) {
		int a, b;
		cin >> a >> b;
		vertexs[a].push_back(b);
		vertexs[b].push_back(a);
	}
	REP(tea, 2) {
		queue<int> next;
		next.push(s);
		while (next.empty() == false) {
			int now = next.front();
			next.pop();
			cnt[tea][dp[tea][now]]++;
			check[tea][dp[tea][now]].push_back(now);
			for (auto x : vertexs[now]) {
				if (dp[tea][x] > dp[tea][now] + 1) {
					dp[tea][x] = dp[tea][now] + 1;
					next.push(x);
				}
			}
		}
		swap(s, t);
	}
	int ans = 0;
	for (int i = 0; i <= dp[0][t] - 2; ++i) {
		ans += cnt[0][i] * cnt[1][dp[0][t] - 2 - i];
		for (auto x : check[0][i]) {
			cnt[1][dp[1][x]]--;
		}
	}
	cout << ans << endl;
}

#undef int

int main() {
	init();
	solve();
}
