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: Accepted)

let split_on_char sep s =
  let open String in
  let r = ref [] in
  let j = ref (length s) in
  for i = length s - 1 downto 0 do
    if get s i = sep then begin
      r := sub s (i + 1) (!j - i - 1) :: !r;
      j := i
    end
  done;
  sub s 0 !j :: !r

let dfs vs n =
  let d = Array.make n (-1) in
  let rec doit i j =
    d.(i) <- j;
    List.iter (fun v -> if d.(v) = (-1) then doit v j) vs.(i) in
  Array.iteri (fun i e -> if e = (-1) then doit i i) d;
  d

let () =
  let read_ints () = read_line () |> split_on_char ' ' |>  List.map int_of_string in
  let (n, m) = match read_ints () with
    | [n; m] -> (n, m)
    | _ -> assert false in
  let vs = Array.make n [] in
  for _ = 0 to m - 1 do
    match read_ints () with
    | [s; t] ->
      vs.(s) <- t :: vs.(s);
      vs.(t) <- s :: vs.(t);
    | _ -> assert false
  done;
  let d = dfs vs n in
  let q = read_int () in
  for _ = 0 to q - 1 do
    match read_ints () with
    | [s; t] -> Printf.printf (if d.(s) = d.(t) then "yes\n" else "no\n")
    | _ -> assert false
  done