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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define repp(i, m, n) for(int i = m; i < n; i++)
#define reps(i, n) for(int i = 1; i <= n; i++)
const int    INF = 100000000;
const double EPS = 1e-10;
const int    MOD = 1000000007;
using namespace std;
typedef long long ll;
typedef pair<int, int> pai;
typedef pair<ll,ll> pal;

int skara[100010];
int sdone[100010];
int tkara[100010];
int tdone[100010];
vector<int> edge[100010];
queue<int> q1, q2;

void bfss(int v)
{
	sdone[v]=1;
	q1.push(v);
	while(!q1.empty())
	{
		int u=q1.front();
		q1.pop();
		rep(i, edge[u].size())
		{
			if(sdone[edge[u][i]]==0)
			{
				skara[edge[u][i]]=skara[u]+1;
				sdone[edge[u][i]]=1;
				q1.push(edge[u][i]);
			}
		}
	}
}

void bfst(int v)
{
	tdone[v]=1;
	q2.push(v);
	while(!q2.empty())
	{
		int u=q2.front();
		q2.pop();
		rep(i, edge[u].size())
		{
			if(tdone[edge[u][i]]==0)
			{
				tkara[edge[u][i]]=tkara[u]+1;
				tdone[edge[u][i]]=1;
				q2.push(edge[u][i]);
			}
		}
	}
}

int main()
{
	int n, m, s, t, x, y;
	cin >> n >> m >> s >> t;
	rep(i, m)
	{
		cin >> x >> y;
		edge[x].push_back(y);
		edge[y].push_back(x);
	}
	bfss(s);
	bfst(t);
	int kyori=skara[t];
	kyori-=2;
	int ss[100010]={}, tt[100010]={};
	reps(i, n)
	{
		ss[skara[i]]++;
		tt[tkara[i]]++;
	}
	ll ans=0;
	if(skara[t]>2)
	{
		ans+=(ss[kyori]+tt[kyori]);
		reps(j, kyori-1) ans+=((ll)ss[j])*((ll)tt[kyori-j]);
	}
	else if(skara[t]==2) ans=1;
	cout << ans << endl;
	return 0;
}