//想定WA解法(x回目の交換だけ飛ばして行い、配列Cとのdiffを取る)
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;

int n, k, x;
int a[100001], b[100001];
int c[100001];
int d[100001];

int main()
{
	int i;
	cin >> n >> k >> x;
	for( i = 1; i <= k; i++ ){
		if( i == x ){
			string yuki, coder;
			cin >> yuki >> coder;
		}
		else{
			cin >> a[i] >> b[i];
		}
	}
	for( i = 1; i <= n; i++ ){
		cin >> c[i];
	}
	
	for( i = 1; i <= n; i++ ){
		d[i] = i;
	}
	
	for( i = 1; i <= k; i++ ){
		if( i != x ){
			swap(d[a[i]], d[b[i]]);
		}
	}
	
	vector<int> out;
	for( i = 1; i <= n; i++ ){
		if( c[i] != d[i] ){
			out.push_back(i);
		}
	}
	cout << out[0] << " " << out[1] << endl;
	return 0;
}