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 <stdio.h>
#include <stdlib.h>
#define MAX 100010
#define INFTY 10000000
#define NIL -1

struct Node {
	int value;
	struct Node *next;
};
typedef struct Node node_t;


int N;
int M;
node_t *G[MAX];
int color[MAX];

node_t* addNode(node_t *node, int value)
{
	node_t *new = (node_t*)malloc(sizeof(node_t));
	new->value = value;
	new->next = node;
	return new;
}

void dfs(int r, int c)
{
	color[r] = c;
	node_t *n;
	for (n=G[r]; n!=NULL; n=n->next) {
		int v = n->value;
		if (color[v] == NIL) {
			color[v] = c;
			dfs(v, c);
		}
	}
}

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


int main()
{
	int i, n, u, v;	
	scanf("%d %d", &N, &M);	
	for (i = 0; i < M; i++) {
		scanf("%d %d", &n, &v);
		G[n] = addNode(G[n], v);
		G[v] = addNode(G[v], n);
	}
	
	assignColor();
	
	int Q;
	scanf("%d", &Q);
	for (i = 0; i < Q; i++) {
		scanf("%d %d", &u, &v);
		if (color[u] == color[v]) {
			puts("yes");			
		}
		else {
			puts("no");
		}
	}

	return 0;
}