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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

void unite(long *affiliation, long *group, long s, long t) {
  long i,j;
  long p = affiliation[s];
  long q = affiliation[t];
  j = 0;
  for (i = 0; j < group[q]; i++) {
    if(affiliation[i] == q) {
      affiliation[i] = p;
      j++;
    }
  }
  group[p] += group[q];
  group[q] = 0;
}

int main()
{
  long n,m,i,s,t,p,q;
  long *affiliation,*group;
  scanf("%ld %ld",&n,&m);
  affiliation = (long*)malloc(sizeof(long) * n);
  for (i = 0; i < n; i++) {
    affiliation[i] = -1;
  }
  group = (long*)malloc(sizeof(long) * m);
  for (i = 0; i < m; i++) {
    group[i] = 0;
  }
  p = 0;
  for (i = 0; i < m; i++) {
    scanf("%ld %ld", &s, &t);
    if(s != t) {
      if(affiliation[s] == -1 && affiliation[t] == -1) {
	affiliation[s] = p;
	affiliation[t] = p;
	group[p] += 2;
	p++;
      }
      else if(affiliation[s] != -1 && affiliation[t] == -1) {
	affiliation[t] = affiliation[s];
	group[affiliation[s]]++;
      }
      else if(affiliation[s] == -1 && affiliation[t] != -1) {
	affiliation[s] = affiliation[t];
	group[affiliation[t]]++;
      }
      else if(affiliation[s] != affiliation[t]){
	unite(affiliation,group,s,t);
      }
    }
  }
  scanf("%ld",&q);
  for (i = 0; i < q; i++) {
    scanf("%ld %ld", &s, &t);
    if(affiliation[s] == affiliation[t] && affiliation[s] != -1) printf("yes\n");
    else printf("no\n");
  }
  free(affiliation);
  free(group);
  
  return 0;
}