F - Find The Bone

F - Find The Bone

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 hasm 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.

Example

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

题意： 第一行输入整数n，m，k。表示有n个杯子，从1到n编号，沿x轴放置在一个有m个孔的桌子上；
　　　 第二行输入m个数，表示该位置有孔；
　　　 第三行k次操作，每次输入两个数x，y。交换x，y位置所在的杯子里的东西。骨头放在1号杯子中，如果骨头所在的位置有孔，
　　　 则骨头会掉进去（骨头就移动不了了）。
　　　 输出骨头的最终位置。

题解：1.用0表示该位置没孔，1表示该位置有孔，用ans标记骨头的位置，
　　　2.每次操作，先判断骨头的位置有没有孔，有孔，则骨头的位置确定。
　　  3.没有孔，就判断该两个位置上有没有骨头，有骨头，就更新骨头的位置。
　　　4.再次判断骨头的位置有没有孔，有孔，则骨头的位置确定。

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstdio>
using namespace std;
int a[1000005];
int main()
{
    int i,n,m,k,x,y,t,ans=1;
    scanf("%d %d %d",&n,&m,&k);

    for(i=1;i<=n;i++)
        a[i]=0;

    for(i=0;i<m;i++)
    {
        scanf("%d",&x);
        a[x]=1;
    }

    for(i=0;i<k;i++)
    {
        scanf("%d %d",&x,&y);
        if(a[ans]==1)
            break;
        if(x==ans)
            ans=y;
        else if(y==ans)
            ans=x;
        if(a[ans]==1)
            break;
    }
    cout<<ans<<endl;
}