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 <queue>
#include <cstdio>
#include <vector>
 
#define INF (1 << 23)
 
using namespace std;

void reader(int *x)
{
    int k = getchar_unlocked(); *x = k - '0';
    
    while(true)
    {
        k = getchar_unlocked();
        
        if(k < '0' || k > '9') break;
        
        *x = ((*x) << 1) + ((*x) << 3) + k - '0';
    }
}

char memarr[17000000]; void *mem = memarr;
 
int main()
{
    int N, M, s, t, u, v;
 
    reader(&N);
    reader(&M);
    reader(&s); s--;
    reader(&t); t--;
 
    vector<vector<int> > G(N);
 
    for (int i = 0; i < M; i++)
    {
        reader(&u);
        reader(&v);
 
        G[u - 1].push_back(v - 1);
        G[v - 1].push_back(u - 1);
    }
 
    vector<int> d1(N, INF); d1[s] = 0;
    vector<int> d2(N, INF); d2[t] = 0;
 
    queue<int> que;
 
    que.push(s);
 
    while (!que.empty())
    {
        int node = que.front(); que.pop();
 
        for (int i = 0; i < G[node].size(); i++)
        {
            if (d1[G[node][i]] == INF)
            {
                d1[G[node][i]] = d1[node] + 1; que.push(G[node][i]);
            }
        }
    }
 
    que.push(t);
 
    while (!que.empty())
    {
        int node = que.front(); que.pop();
 
        for (int i = 0; i < G[node].size(); i++)
        {
            if (d2[G[node][i]] == INF)
            {
                d2[G[node][i]] = d2[node] + 1; que.push(G[node][i]);
            }
        }
    }
 
    int shortest = d1[t];
 
    vector<int> r1(shortest - 1, 0); for (int i = 0; i < N; i++) { if (d1[i] < shortest - 1) r1[d1[i]]++; }
    vector<int> r2(shortest - 1, 0); for (int i = 0; i < N; i++) { if (d2[i] < shortest - 1) r2[d2[i]]++; }
 
    long long ret = 0;
 
    for (int i = 0; i < shortest - 1; i++) ret += 1LL * r1[i] * r2[shortest - i - 2];
     
    printf("%lld\n", ret);
 
    return 0;
}