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<stack>
#include<vector>

using namespace std;

#define MAX 100000
#define NIL -1

int n;
int color[MAX];
vector<int> G[MAX];


void dfs(int u, int id){
  
  stack<int> S;
  S.push(u);
  color[u] = id;

  int present;
  while( !S.empty() ){
    present = S.top();
    S.pop();

    for(int i=0;i<G[present].size();i++){
      int v = G[present][i];
      if( color[v] == NIL ){
        color[v] = id;
        S.push(v);
      }    
    }  
  }
}

void assignColor(){
 
  for(int i=0;i<n;i++){
    color[i] = NIL;
  } 

  int id = 0;
  for(int u=0;u<n;u++){
    if(color[u] == NIL){
      dfs(u,id);
      id = id + 1;
    }
  }
}

int main(void){

  int m;
  cin >> n >>  m;

  int u,v;
  for(int i=0;i<m;i++){
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }

  assignColor();

  int q;
  cin >> q;

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

  return 0;
}