/*******************************************************
 * Copyright (C) 2015 Haotian Wu
 *
 * This file is the solution to the question:
 * https://www.hackerrank.com/challenges/rust-murderer
 *
 * Redistribution and use in source and binary forms are permitted.
 *******************************************************/
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <map>
#include <set>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// We use BFS here. Since the graph is huge but sparse, we can add our custom exit condition.
map<int,int> adj[200001];
int visited[200001],q[200001],dis[200001];
int main()
{
    int tt;
    scanf("%d",&tt);
    while (tt--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        for (int i=0;i<=n;i++)
            adj[i].clear();
        for (int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            adj[a][b]=1;
            adj[b][a]=1;
        }
        int s;
        scanf("%d",&s);
        int be=0,ed=1;
        memset(visited,0,sizeof(visited));
        q[0] = s;
        visited[s] = 1;
        dis[s] = 0;
        while (be<ed)
        {
            for (int i=1;i<=n;i++)
                if (visited[i] == 0 && adj[q[be]][i] != 1)
                {
                    q[ed++] = i;
                    visited[i] = 1;
                    dis[i] = dis[q[be]] + 1;
                    //printf("Dis of %d is %d \n",i,dis[i]);
                }
            if (ed>=n)
                break;
            be++;
        }
        for (int i=1;i<=n;i++)
            if (i!=s)
                printf("%d ",dis[i]);
        printf("\n");
    }
}