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: Accepted)

#include <cassert>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <bitset>
#include <complex>

#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#define rep(i,n) for(int i=0;i<(n); i++)
#define rep2(i,n) for(int i=1;i<=(n);i++)
#define WHITE 0
#define GRAY 1
#define BLACK 2

using namespace std;
typedef long long ll;
const int NMAX = 100000;
const int k = 200000;
static const int INF = 1<<21;
static const int NIL = -1;

int dp[3001][3001];
ll gcd(ll a, ll b){
if(a<b) swap(a,b);
    if(b==0) return a;
    return gcd(b, a%b);
}

int n,color[NMAX];
vector<int> G[NMAX];

void dfs(int r,int c){
    stack<int> S;
    S.push(r);
    color[r] = c;
    while(!S.empty()){
        int u=S.top();
        S.pop();
        rep(i,G[u].size()){
            int v=G[u][i];
            if(color[v]==NIL){
                color[v]=c;
                S.push(v);
            }
        }
    }
}

void assignColor(){
    int id=1;
    rep(i,n) color[i]=NIL;
    rep(u,n){
        if(color[u]==NIL){
            dfs(u,id++);
        }
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(0); //この2行はより速度を向上させる
    // 追加２, 以降 cin の入力元が 'input.txt' になる


    int m,s,t,q;
    cin >> n >> m;
    rep(i,m){
        cin >> s >> t;
        G[s].push_back(t);
        G[t].push_back(s); //無向グラフなので双方に格納する
    }
    assignColor(); //色塗り

    cin >> q;

    rep(i,q){
        cin >> s >> t;
        if(color[s]==color[t]){
            cout << "yes" << endl;
        }else{
            cout << "no" << endl;
        }
    }
    return 0;
}