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++11 (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <queue>
#include <stack>
#include <map>
#define FOR(i,b,e) for(int i=b;i<e;i++)
#define REP(i,n) FOR(i,0,n)
using namespace std;

const int N = 10000;
int n, m, q;
int M[N][N];
bool V[N];

int bfs(int u, int v) {
  int s;
  queue<int> q;
  REP(i, n) V[i] = false;

  s = u;
  q.push(s);
  V[s] = true;

  while (!q.empty() && !V[v]) {
    s = q.front(); q.pop();
    REP(i, n) {
      if (M[s][i] == 1 && !V[i]) {
        q.push(i);
        V[i] = true;
      }
    }
  }

  return V[v];
}

int main() {
	cin.tie(0);
 	ios::sync_with_stdio(false);

  REP(i, n) {
    REP(j, n) {
      M[i][j] = 0;
    }
  }

  cin >> n >> m;
  REP(i, m) {
    int u, v;
    cin >> u >> v;
    M[u][v] = M[v][u] = 1;
  }

  cin >> q;
  REP(i, q) {
    int u, v;
    string answer;
    cin >> u >> v;
    answer = bfs(u, v) ? "yes" : "no";
    cout << answer << endl;
  }
}