Alice and Bob are playing a game. This game is played with several identical pots and one marble. When the game starts, Alice puts the pots in one line and puts the marble in one of the pots. After that, Bob cannot see the inside of the pots. Then Alice makes a sequence of swappings and Bob guesses which pot the marble is in. In each of the swapping, Alice chooses two different pots and swaps their positions.

Unfortunately, Alice's actions are very fast, so Bob can only catch k ofm swappings and regard these k swappings as all actions Alice has performed. Now given the initial pot the marble is in, and the sequence of swappings, you are asked to calculate which pot Bob most possibly guesses. You can assume that Bob missed any of the swappings with equal possibility.

Input

There are several test cases in the input file. The first line of the input file contains an integerN (N ≈ 100), then N cases follow.

The first line of each test case contains 4 integers n, m, k and s(0 < s ≤ n ≤ 50, 0 ≤ k ≤ m ≤ 50), which are the number of pots, the number of swappings Alice makes, the number of swappings Bob catches and index of the initial pot the marble is in. Pots are indexed from 1 to n

Outout

For each test case, output the pot that Bob most possibly guesses. If there is a tie, output the smallest one.

Sample Input

3
3 1 1 1
1 2
3 1 0 1
1 2
3 3 2 2
2 3
3 2
1 2

Sample Output

2
1
3

題目大意：N個容器，M次兩兩交換，其中K次是可以知道的，一開始珠子放在其中一個容器裡，問你交換完以後，珠子在哪個容器的概率最大

思路：用DP[m][k][n] 表示 m次交換，知道了其中的k次，結尾為n的方案數

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using  namespace std;
typedef long long ll;
const  int maxn = 60;
int n,m,k,s;
int A[maxn],B[maxn];
ll dp[maxn][maxn][maxn];
int main(){

    int ncase;
    cin >> ncase;
    while(ncase--){
        scanf("%d%d%d%d",&n,&m,&k,&s);
        for(int i = 1; i <= m; i++){
            scanf("%d%d",&A[i],&B[i]);
        }
        memset(dp,0,sizeof dp);
        dp[0][0][s] = 1;
        for(int i = 1; i <= m; i++){
            dp[i][0][s] = 1;
            for(int j = 1; j <= i&&j <= k; j++){
                dp[i][j][A[i]] = dp[i-1][j-1][B[i]];
                dp[i][j][B[i]] = dp[i-1][j-1][A[i]];
                for(int d = 1; d <= n; d++){
                    dp[i][j][d] += dp[i-1][j][d];
                    if(d!=A[i]&&d!=B[i]){
                        dp[i][j][d] += dp[i-1][j-1][d];
                    }
                }
            }
        }
        int idx = 1;
        for(int i = 2; i <= n; i++){
            if(dp[m][k][i] > dp[m][k][idx]){
                idx = i;
            }
        }
        cout<<idx<<endl;
    }
    return 0;
}