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++14 (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<stack>
using namespace std;

const int N = 100000;
const int NIL = -1;

int n;
vector<int> G[N];
int color[N];

void dfs(int r, int c){
    stack<int> S;
    S.push(r);
    color[r] = c;

    while(!S.empty()){
        int x = S.top(); S.pop();
        for(int i = 0; i < G[x].size(); i++){
            int v = G[x][i];
            if(color[v] == NIL){
                color[v] = c;
                S.push(v);
            }
        }
    }
}

void solve(){
    int id = 1;
    for(int i = 0; i < n; i++)  color[i] = NIL;
    for(int u = 0; u < n; u++){
        if(color[u] == NIL) dfs(u, id++);
    }
}

int main(){
    int q, s, t;
    cin >> n >> q;

    while(q-- > 0){
        cin >> s >> t;
        G[s].push_back(t);
        G[t].push_back(s);
    }

    solve();

    cin >> q;
    while(q-- > 0){
        cin >> s >> t;
        if(color[s] == color[t])    cout << "yes" << endl;
        else                        cout << "no" << endl;
    }

    return 0;
}