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<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int,int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-11;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;
struct edge { int to, cost; };
int V;
vector<edge> G[100000];
int d[100000]; int ds[100000]; int dg[100000]; ll cnt[100000] = {};
void dijkstra(int s) {
	priority_queue<P, vector<P>, greater<P>>que;
	fill(d, d + V, (int)MOD);
	d[s] = 0;
	que.push(P(0, s));
	while (!que.empty()) {
		P p = que.top(); que.pop();
		int v = p.second;
		if (d[v] < p.first)continue;
		for (int i = 0; i < (int)G[v].size(); i++) {
			edge e = G[v][i];
			if (d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
}
int main() {
	int m;
	cin >> V >> m;
	int s, g; cin >> s >> g; s--; g--;
	rep(i, m) {
		int x, y; cin >> x >> y; x--; y--;
		G[x].push_back({ y,1 });
		G[y].push_back({ x,1 });
	}
	dijkstra(s);
	rep(i, V) {
		ds[i] = d[i];
	}
	dijkstra(g);
	rep(i, V) {
		dg[i] = d[i];
	}
	int r = ds[g];
	rep(i, V) {
		if (ds[i] < (int)MOD) {
			cnt[ds[i]]++;
		}
	}
	ll res = 0;
	rep(i, V) {
		int x = r - dg[i] - 2;
		if (x < 0)continue;
		res += cnt[x];
	}
	cout << res << endl;
	return 0;
}
