There are n cities in Berland and some pairs of them are connected by two-way roads. It is guaranteed that you can pass from any city to any other, moving along the roads. Cities are numerated from 1 to n.

Two fairs are currently taking place in Berland — they are held in two different cities a and b (1≤a,b≤n; a≠b).

Find the number of pairs of cities x and y (x≠a,x≠b,y≠a,y≠b) such that if you go from x to y you will have to go through both fairs (the order of visits doesn’t matter). Formally, you need to find the number of pairs of cities x,y such that any path from x to y goes through a and b (in any order).

Print the required number of pairs. The order of two cities in a pair does not matter, that is, the pairs (x,y) and (y,x) must be taken into account only once.

Input
The first line of the input contains an integer t (1≤t≤4⋅104) — the number of test cases in the input. Next, t test cases are specified.

The first line of each test case contains four integers n, m, a and b (4≤n≤2⋅105, n−1≤m≤5⋅105, 1≤a,b≤n, a≠b) — numbers of cities and roads in Berland and numbers of two cities where fairs are held, respectively.

The following m lines contain descriptions of roads between cities. Each of road description contains a pair of integers ui,vi (1≤ui,vi≤n, ui≠vi) — numbers of cities connected by the road.

Each road is bi-directional and connects two different cities. It is guaranteed that from any city you can pass to any other by roads. There can be more than one road between a pair of cities.

The sum of the values of n for all sets of input data in the test does not exceed 2⋅105. The sum of the values of m for all sets of input data in the test does not exceed 5⋅105.

Output
Print t integers — the answers to the given test cases in the order they are written in the input.

Example
inputCopy
3
7 7 3 5
1 2
2 3
3 4
4 5
5 6
6 7
7 5
4 5 2 3
1 2
2 3
3 4
4 1
4 2
4 3 2 1
1 2
2 3
4 1
outputCopy
4
0
1

题目大意：找出一些点对，满足一个点到另一个点一定会经过a,b两个点。输出满足条件的点对数目。

所以我们就是要找到必须通过a才能到b的点的个数，和必须通过b才能到a的点的个数。然后相乘即可。

怎么求呢？必须通过a才能到b，实际上我们对b点跑bfs，然后并且限制不能经过a点，那么看有多少个不能到达的点即可。后者一样。

AC代码：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+10;
int T,n,m,a,b,res,vis[N],cnt;
vector<int> v[N];
inline void add(int a,int b){v[a].push_back(b); v[b].push_back(a);}
inline int bfs(int x,int y){
    queue<int> q;   q.push(x);  cnt=2;  memset(vis,0,8*(n+1)); vis[x]=vis[y]=1;
    while(q.size()){
        int u=q.front();    q.pop();
        for(int i=0,to;i<v[u].size();i++){
            to=v[u][i];
            if(!vis[to])    vis[to]=1,q.push(to),cnt++;
        }
    }
    return n-cnt;
}
signed main(){
    ios::sync_with_stdio(false);  cin.tie(nullptr); cout.tie(nullptr);
    cin>>T;
    while(T--){
        cin>>n>>m>>a>>b;
        for(int i=1,x,y;i<=m;i++)   cin>>x>>y,add(x,y);
        res=bfs(a,b); 	res*=bfs(b,a);
        cout<<res<<'\n';
        for(int i=1;i<=n;i++)   v[i].clear();
    }
    return 0;
}