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;

#define rep(i,a,b) for(int i=a;i<b;i++)
#define INF INT_MAX/2
typedef long long ll;
typedef vector<ll> vi;
typedef vector<bool> vb;
typedef pair<int, ll> pii;

struct Comp {
	bool operator() (pii a, pii b) {
		return a.second > b.second;
	}
};

int main()
{
	int N, M, s, t;
	cin >> N >> M >> s >> t;
	s--; t--;

	map< int, vector<pii> > edge;
	rep(i, 0, M)
	{
		int x, y;
		cin >> x >> y;
		x--; y--;
		edge[x].push_back(pii(y, 1));
		edge[y].push_back(pii(x, 1));
	}

	vi dist(N, INF);
	vb done(N, false);
	priority_queue<pii, vector<pii>, Comp> que;
	dist[s] = 0;
	que.push(pii(s, 0));
	while (!que.empty())
	{
		pii p = que.top(); que.pop();
		if (done[p.first]) continue;
		done[p.first] = true;
		for (pii pp : edge[p.first])
		{
			int newcost = p.second + pp.second;
			if (newcost < dist[pp.first])
			{
				dist[pp.first] = newcost;
				que.push(pii(pp.first, newcost));
			}
		}
	}

	vi dist2(N, INF);
	vb done2(N, false);
	dist2[t] = 0;
	que.push(pii(t, 0));
	while (!que.empty())
	{
		pii p = que.top(); que.pop();
		if (done2[p.first]) continue;
		done2[p.first] = true;
		for (pii pp : edge[p.first])
		{
			int newcost = p.second + pp.second;
			if (newcost < dist2[pp.first])
			{
				dist2[pp.first] = newcost;
				que.push(pii(pp.first, newcost));
			}
		}
	}

	int len = dist[t];
	vi from_s(len + 1, 0);
	vi from_t(len + 1, 0);
	rep(i, 0, N)
	{
		if (dist[i] <= len)
			from_s[dist[i]]++;
		if (dist2[i] <= len)
			from_t[dist2[i]]++;
	}

	ll ans = 0;
	rep(i, 0, len-1)
	{
		ans += from_s[i] * from_t[len - 1 - i - 1];
	}
	cout << ans << endl;
}