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 Julia (1.4.0) (judge verdict: Wrong Answer)

# SPDX-License-Identifier: X11
# 2020-08-30
# Coprime (500pt)

function main()
  N = parse(Int, readline())
  A = parse.(Int, split(readline(), " "))

  if reduce(gcd, A) > 1
    println("not coprime")
    return
  end

  sort!(A)
  bound = (Int ∘ trunc ∘ sqrt)(A[N])

  for i ∈ 1:(N - 1)
    if A[i] == A[i + 1]
      println("setwise coprime")
      return
    end
  end

  if bound < 2
    println("pairwise coprime")
    return
  end

  ps = erato(bound)
  as::Vector{Bool} = falses(bound)

  for i ∈ A
    if hash!(as, i, ps) === nothing
      println("setwise coprime")
      return
    end
  end

  println("pairwise coprime")
end

function erato(bound)::Vector{Bool}
  A = falses(bound)

  bound < 2 && return A

  A[2] = true

  for i ∈ 3:2:bound
    A[i] = true
  end

  for i ∈ 3:2:bound
    if A[i]
      for j ∈ i:2:(bound ÷ i)
        A[i * j] = false
      end
    end
  end

  return A
end

function hash!(as, n, ps)
  if n & 1 == 0
    if as[2]
      return nothing
    else
      as[2] = true
    end
  end

  for i ∈ 3:2:(Int ∘ trunc ∘ sqrt)(n)
    if ps[i] && n % i == 0
      if as[i]
        return nothing
      else
        as[i] == true
      end
    end
  end

  return n
end

main()