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 $b$ ($1 \le a, b \le n$; $a \ne b$).
Find the number of pairs of cities $x$ and $y$ ($x \ne a, x \ne b, y \ne a, y \ne b$) 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 $t$ ($1 \le t \le 4\cdot10^4$) — 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 $b$ ($4 \le n \le 2\cdot10^5$, $n - 1 \le m \le 5\cdot10^5$, $1 \le a,b \le n$, $a \ne b$) — 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 $u_i, v_i$ ($1 \le u_i, v_i \le n$, $u_i \ne v_i$) — 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 $2\cdot10^5$. The sum of the values of $m$ for all sets of input data in the test does not exceed $5\cdot10^5$.
-----Output-----
Print $t$ integers — the answers to the given test cases in the order they are written in the input.
-----Example-----
Input
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
Output
4
0
1
import sys
readline = sys.stdin.readline

def dfs(s, t, Edge):
    N = len(Edge)
    used = [0]*N
    used[s] = 1
    used[t] = 1
    stack = [s]
    while stack:
        vn = stack.pop()
        for vf in Edge[vn]:
            if not used[vf]:
                used[vf] = 1
                stack.append(vf)
    return used

T = int(readline())
Ans = [None]*T
for qu in range(T): 
    N, M, A, B = map(int, readline().split())
    A -= 1
    B -= 1
    Edge = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(int, readline().split())
        a -= 1
        b -= 1
        Edge[a].append(b)
        Edge[b].append(a)
    sa = dfs(A, B, Edge)
    sb = dfs(B, A, Edge)
    ca = 0
    cb = 0
    for fa, fb in zip(sa, sb):
        if fa == fb:
            continue
        if fa:
            ca += 1
        elif fb:
            cb += 1
    Ans[qu] = ca*cb
print('\n'.join(map(str, Ans)))
import sys
readline = sys.stdin.readline

def dfs(s, t, Edge):
    N = len(Edge)
    used = [0]*N
    used[s] = 1
    used[t] = 1
    stack = [s]
    while stack:
        vn = stack.pop()
        for vf in Edge[vn]:
            if not used[vf]:
                used[vf] = 1
                stack.append(vf)
    return used

T = int(readline())
Ans = [None]*T
for qu in range(T): 
    N, M, A, B = map(int, readline().split())
    A -= 1
    B -= 1
    Edge = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(int, readline().split())
        a -= 1
        b -= 1
        Edge[a].append(b)
        Edge[b].append(a)
    sa = dfs(A, B, Edge)
    sb = dfs(B, A, Edge)
    ca = 0
    cb = 0
    for fa, fb in zip(sa, sb):
        if fa == fb:
            continue
        if fa:
            ca += 1
        elif fb:
            cb += 1
    Ans[qu] = ca*cb
print('\n'.join(map(str, Ans)))