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: Memory Limit Exceeded)

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


typedef struct {
  unsigned char a0: 1;
  unsigned char a1: 1;
  unsigned char a2: 1;
  unsigned char a3: 1;
  unsigned char a4: 1;
  unsigned char a5: 1;
  unsigned char a6: 1;
  unsigned char a7: 1;
}boolean;

void set_boolean(boolean *bool, int ad, char in) {
  switch (ad) {
  case 0:
    bool->a0 = in;
    break;
  case 1:
    bool->a1 = in;
    break;
  case 2:
    bool->a2 = in;
    break;
  case 3:
    bool->a3 = in;
    break;
  case 4:
    bool->a4 = in;
    break;
  case 5:
    bool->a5 = in;
    break;
  case 6:
    bool->a6 = in;
    break;
  case 7:
    bool->a7 = in;
    break;
  default:
    break;
  }
}

char check_boolean(boolean *bool, int ad) {
  switch (ad) {
  case 0:
    return bool->a0;
    break;
  case 1:
    return bool->a1;
    break;
  case 2:
    return bool->a2;
    break;
  case 3:
    return bool->a3;
    break;
  case 4:
    return bool->a4;
    break;
  case 5:
    return bool->a5;
    break;
  case 6:
    return bool->a6;
    break;
  case 7:
    return bool->a7;
    break;
  default:
    printf("miss");
    return 0;
    break;
  }
}

boolean **build_matrix(long n) {
  boolean **a = (boolean**)malloc(sizeof(boolean*) * n);
  long i;
  for (i = 0; i < n; i++) {
    a[i] = (boolean*)calloc(n/8+1,sizeof(boolean));
  }
  return a;
}


void search(boolean **a, long n, boolean **group, long g_num, long u, long *affiliation) {
  if(check_boolean(group[g_num]+u/8,u%8) != 0) return;
  set_boolean(group[g_num]+u/8,u%8,1);
  affiliation[u] = g_num;
  long i;
  for (i = 0; i < n; i++) {
    if(check_boolean(a[u]+i/8,i%8) == 1) {
      search(a,n,group,g_num,i,affiliation);
    }
  }
}

int main()
{
  long i,j,n,m,u,v;
  boolean **a;
  scanf("%ld %ld",&n,&m);
  a = build_matrix(n);

  for (i = 0; i < m; i++) {
    scanf("%ld",&u);
    scanf("%ld",&v);
    set_boolean(a[u]+v/8,v%8,1);
    set_boolean(a[v]+u/8,u%8,1);
  }

  boolean **group;
  group = (boolean**)malloc(sizeof(boolean*) * n);
  long *affiliation;
  affiliation = (long*)malloc(sizeof(long) * n);
  for (i = 0; i < n; i++) {
    affiliation[i] = -1;
  }

  j = 0;
  for (i = 0; i < n; i++) {
    if(affiliation[i] == -1) {
      group[j] = (boolean*)calloc(n/8+1,sizeof(boolean));
      search(a,n,group,j,i,affiliation);
      j++;
    }
  }

  long q,s,t;
  scanf("%ld",&q);
  for (i = 0; i < q; i++) {
    scanf("%ld %ld",&s,&t);
    if(check_boolean(group[affiliation[s]]+t/8,t%8) == 1) printf("yes\n");
    else printf("no\n");

  }



  for (i = 0; i < n; i++) {
    free(a[i]);
  }
  for (i = 0; i < j; i++) {
    free(group[i]);
  }

  free(a);
  free(group);
  free(affiliation);
  return 0;
}