Problem:
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Connected Components

Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.

Input

In the first line, two integer $n$ and $m$ are given. $n$ is the number of users in the SNS and $m$ is the number of relations in the SNS. The users in the SNS are identified by IDs $0, 1, ..., n-1$.

In the following $m$ lines, the relations are given. Each relation is given by two integers $s$ and $t$ that represents $s$ and $t$ are friends (and reachable each other).

In the next line, the number of queries $q$ is given. In the following $q$ lines, $q$ queries are given respectively. Each query consists of two integers $s$ and $t$ separated by a space character.

Output

For each query, print "yes" if $t$ is reachable from $s$ through the social network, "no" otherwise.

Constraints

$2 \leq n \leq 100,000$

$0 \leq m \leq 100,000$

$1 \leq q \leq 10,000$

Sample Input

10 9
0 1
0 2
3 4
5 7
5 6
6 7
6 8
7 8
8 9
3
0 1
5 9
1 3

Sample Output

yes
yes
no

Solution in C# (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Practice
{
    class Program
    {
        const int WHITE = -10;
        const int GRAY = -9;
        //const int BLACK = 2;
        static void Main(string[] args)
        {
            SolveA(args);
            //SolveB(args);
            //SolveC(args);
            //SolveD(args);


        }
        static void SolveA(string[] args)
        {
            string[] szarr = Console.ReadLine().Split(' ');
            int n = int.Parse(szarr[0]);
            int m = int.Parse(szarr[1]);
            List<int>[] g = new List<int>[n];
            for(int i = 0; i < n; ++i)
            {
                g[i] = new List<int>();
            }
            for(int i=0;i<m;++i)
            {
                szarr = Console.ReadLine().Split(' ');
                int v0 = int.Parse(szarr[0]);
                int v1 = int.Parse(szarr[1]);
               
                g[v0].Add(v1);
                g[v1].Add(v0);
            }
            int q = int.Parse(Console.ReadLine());

            int[] color = new int[n];
            int[] d = new int[n];
            var queue = new Queue<int>();

            for (int i=0;i<q;++i)
            {
                szarr = Console.ReadLine().Split(' ');
                int v0 = int.Parse(szarr[0]);
                int v1 = int.Parse(szarr[1]);
                // ２人はフレンド？
                for (int j = 0; j < color.Length; ++j) { color[j] = WHITE; }
                for (int j = 0; j < d.Length; ++j) { d[j] = -1; }
                queue.Clear();
                bool b = BFS(v0,v1, color, d, queue, g);

                if(b)
                {
                    Console.WriteLine("yes");
                }
                else
                {
                    Console.WriteLine("no");
                }
            }


        }
        static bool BFS(int s, int s0, int[] color, int[] d, Queue<int> q, List<int>[] g)
        {
            color[s] = GRAY;
            d[s] = 0;
            q.Enqueue(s);
            while(q.Count >= 1)
            {
                int u = q.Dequeue();
                for(int i = 0;i<g[u].Count;++i)
                {
                    int v = g[u][i];
                    if( v == s0) { return true; }
                    if(color[v] == WHITE)
                    {
                        color[v] = GRAY;
                        d[v] = d[u] + 1;
                        q.Enqueue(v);
                    }
                }
                color[u] = s;
            }
            return false;
         }
    }

}