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<iostream>
#include<vector>
using namespace std;
static const int MAX = 100000;
static const int NIL = -1;

int n;
vector<int> G[MAX]; // List of vectors
int color[MAX];

void dfs_visit(int v, int id) {
    color[v] = id;
    for (int i=0; i<G[v].size(); i++) {
        int u = G[v][i];
        if (color[u]==NIL) {
            dfs_visit(u, id);
        }
    }
}

void dfs() {
    for (int v=0; v<n; v++) {
        color[v] = NIL;
    }
    int id = 0; // Cluster id
    for (int v=0; v<n; v++) {
        if (color[v]==NIL) {
            // If not clustered yet
            dfs_visit(v, id++);
        }
    }
}

int main() {
    int s,t,m,q;
    cin >> n >> m;
    for (int i=0; i<m; i++) {
        cin >> s >> t;
        G[s].push_back(t);
        G[t].push_back(s);
    }

    dfs();

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

 return 0;
}