796B Find The Bone

B. Find The Bone

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Zane the wizard is going to perform a magic show shuffling the cups.

There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.

The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = u_i and x = v_i. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.

Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.

Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 10⁶, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·10⁵) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.

The second line contains m distinct integers h₁, h₂, ..., h_m (1 ≤ h_i ≤ n) — the positions along the x-axis where there is a hole on the table.

Each of the next k lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — the positions of the cups to be swapped.

Output

Print one integer — the final position along the x-axis of the bone.

Examples

Input

7 3 4
3 4 6
1 2
2 5
5 7
7 1

Output

1

Input

5 1 2
2
1 2
2 4

Output

2

Note

In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.

In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.

 

 题意：给你n个杯子标号为（1~n），有m个杯子有洞，有k次操作，

          每次操作交换编号为u和编号为v的杯子，如果这个杯子 有洞，骨头就会掉下去，

          初始骨头在位置1，现在问你k此操作以后骨头在哪里

 

 解析：标记洞编号，每次交换前判断骨头是否掉入洞中，

         如果交换的两个杯子中没有骨头，骨头位置没有变，不用处理

         如果交换的两个杯子中有骨头，在 ui 中，并且骨头未掉入洞中， 骨头位置变化到 vi 中

                                                在 vi 中，并且骨头未掉入洞中， 骨头位置变换到 ui 中

        重复 K 次操作，最后找到 骨头的位置 

 

另：用cin cout 可能会超时，较大数据用scanf(); printf();

#include <iostream>
#include <cstdio>
using namespace std;
const int M = 1e6 +10;
int holes[M];
int main(){
    int n, m, k;
    scanf("%d%d%d",&n,&m,&k);
    //cin >> n >> m >> k;
    int stone = 1;
    for(int i = 0; i < m; i++){
        int hol;
        //cin >. hol;
        scanf("%d",&hol);
        holes[hol] ++;
    }
    for(int i = 0; i < k; i++){
        int ui, vi;
        //cin >> ui >> vi;
        scanf("%d%d",&ui,&vi);
        if(ui == stone && holes[ui] != 1){
            stone = vi;
        }
        else if(vi == stone && holes[vi] != 1){
            stone = ui;
        }
    }
    //cout << stone << endl;
    printf("%d",stone);
}