#백준_2644 촌수계산 - Java 자바

# 유형 : 그래프, 탐색, BFS

# 간단한 BFS 문제. 큐에 넣을 때마다 촌수를 1개씩 더해주면서 진행하면 된다.

 

 
import java.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
 
public class p2644 {
    static boolean check = false;
    static int N,M;
    static int start,end;
    static int arr[][];
    static boolean visit[];
    public static void main(String[] args) throws NumberFormatException, IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        N = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        start = Integer.parseInt(st.nextToken());
        end = Integer.parseInt(st.nextToken());
        M = Integer.parseInt(br.readLine());
        
        arr = new int[N+1][N+1];
        visit = new boolean[N+1];
        for(int i=0; i<M; i++) {
            st = new StringTokenizer(br.readLine());
            int u = Integer.parseInt(st.nextToken());
            int v = Integer.parseInt(st.nextToken());
            arr[u][v] = 1;
            arr[v][u] = 1;
        }
        
        bfs(start);
        if(!check) {
            System.out.println(-1);
        }
    }
    
    public static void bfs(int begin) {
        Queue<Point> queue = new LinkedList<Point>();
        visit[begin] = true;
        queue.add(new Point(begin, 0));
        
        while(!queue.isEmpty()) {
            Point po = queue.poll();
            
            if(po.x == end) {
                System.out.println(po.y);
                check = true;
                return;
            }
            
            for(int i=1; i<=N; i++) {
                if(arr[po.x][i]==1 && !visit[i]) {
                    visit[i] = true;
                    queue.add(new Point(i, po.y+1));
                }
            }
        }
    }
}