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 JavaScript (judge verdict: Time Limit Exceeded)

(function main() {
  const lines = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
  const [n, m] = lines.shift().split(' ').map(Number);
  const adjLists = Array.from(Array(n), () => []);
  const d = [];

  let s, t;
  for (let i = 0; i < m; i++) {
    [s, t] = lines.shift().split(' ').map(Number);
    adjLists[s].push(t);
    adjLists[t].push(s);
  }

  function canReach(s, t) {
    const WHITE = 0, GRAY = 1, BLACK = 2;
    const color = Array.from(adjLists, () => WHITE);
    let queue = [s];
    let u, v;
    while ((u = queue.shift()) !== undefined) {
      for (let i = 0; i < adjLists[u].length; i++) {
        v = adjLists[u][i];
        if (v === t) return true;
        if (color[v] === WHITE) {
          color[v] = GRAY;
          queue.push(v);
        }
      }
    }
    return false;
  }

  lines.shift();
  console.log(lines.map(l => canReach(...l.split(' ').map(Number)) ? 'yes' : 'no').join('\n'));
})();