Problem:
Score : 500 points

Problem Statement

We have N integers. The i-th number is A_i.

\{A_i\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.

\{A_i\} is said to be setwise coprime when \{A_i\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.

Determine if \{A_i\} is pairwise coprime, setwise coprime, or neither.

Here, GCD(\ldots) denotes greatest common divisor.

Constraints

2 \leq N \leq 10^6

1 \leq A_i\leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

If \{A_i\} is pairwise coprime, print pairwise coprime; if \{A_i\} is setwise coprime, print setwise coprime; if neither, print not coprime.

Sample Input 1

3
3 4 5

Sample Output 1

pairwise coprime

GCD(3,4)=GCD(3,5)=GCD(4,5)=1, so they are pairwise coprime.

Sample Input 2

3
6 10 15

Sample Output 2

setwise coprime

Since GCD(6,10)=2, they are not pairwise coprime. However, since GCD(6,10,15)=1, they are setwise coprime.

Sample Input 3

3
6 10 16

Sample Output 3

not coprime

GCD(6,10,16)=2, so they are neither pairwise coprime nor setwise coprime.

Solution in OCaml (4.10.0) (judge verdict: Accepted)

let n = read_int ()
let a = Array.init n (fun _ -> Scanf.scanf " %d" (fun i -> i))
let m = Array.fold_left (fun i j -> max i j) 0 a
let check = Array.make (m + 1) false

let d = Array.make (m + 1) 0

let rec pre i =
  if i <= m then (
    if d.(i) = 0 then (
      let rec sieve j =
        let k =  i * j in
        if k <= m then (
          d.(k) <- i;
          sieve (j + 1)
        ) in
      sieve 1;
    );
    pre (i + 1)
  )


let rec solve i =
  if i = n then "pairwise coprime"
  else (
    let rec f a =
      if a > 1 then (
        let p = d.(a) in
        if check.(p) then true
        else (
          check.(p) <- true;
          let rec g a p =
            let r = a mod p in
            if r = 0 then g (a / p) p
            else a in
          f (g a p)
        )
      ) else false in
    if f a.(i) then "setwise coprime"
    else solve (i + 1)
  )

let rec gcd a b =
  let r = a mod b in
  if r = 0 then b
  else gcd b r

let gcd a = Array.fold_left (fun i j -> gcd i j) a.(0) a

let () =
  let g = gcd a in
  let ans =
    if g > 1 then "not coprime"
    else (
      pre 2;
      solve 0
    ) in
  Printf.printf "%s\n" ans