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 OCaml (judge verdict: Time Limit Exceeded)

open Printf
open Scanf

let () =
  let n, m = scanf "%d %d " (fun x y -> (x, y)) in
  let g = Array.make n [] in
  let rec set_graph x =
    if x = 0 then ()
    else begin
        let u, v = scanf "%d %d " (fun x y -> (x, y)) in
        g.(u) <- v :: g.(u);
        g.(v) <- u :: g.(v);
        set_graph (x-1)
      end
  in
  let friend u v =
    let que = Queue.create () in
    let vt = Array.make n true in
    let rec loop () =
      if Queue.is_empty que then false
      else let i = Queue.pop que in
           if i = v then true
           else begin
               List.iter (fun j ->
                          if vt.(j) then begin
                              Queue.push j que; 
                              vt.(j) <- false
                            end) g.(i);
               loop ()
             end
    in
    Queue.push u que;
    vt.(u) <- false;
    loop ()
  in
  let rec iter x =
    if x = 0 then ()
    else begin
        let u, v = scanf "%d %d " (fun x y -> (x, y)) in
        printf "%s\n" (if friend u v then "yes" else "no");
        iter (x-1)
      end
  in
  set_graph m;
  let q = scanf "%d " (fun x -> x) in
  iter q