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 Ruby (2.7.1) (judge verdict: Wrong Answer)

# エラトステネスの篩で素数列挙して高速に素因数分解したい！
# ref: https://atcoder.jp/contests/abc177/editorial/82
class PrimeDivWithSieve
  def initialize(n)
    @sieve = []
    @elements = {}
    (2..n).each do |i|
      next if @elements[i]
      @sieve << i
      @elements[i] = [i]

      sieve_target = i * 2
      while sieve_target <= n do
        if @elements[sieve_target]
          @elements[sieve_target].push(i)
        else
          @elements[sieve_target] = [i]
        end
        sieve_target += i
      end
    end
  end

  def element(num)
    @elements[num] || [num]
  end
end


_n = gets.chomp.split.map(&:to_i)
arr = gets.chomp.split.map(&:to_i)

divisor = PrimeDivWithSieve.new(1_000_000)
common_elms = divisor.element(arr[0])
all_elms = {}
common_elms.each do |i|
  all_elms[i] = true
end

pair = true
set = false
arr[1..-1].each do |i|
  crr_elms = divisor.element(i)
  if pair
    crr_elms.each do |i|
      break pair = false if all_elms[i]
      all_elms[i] = true
    end
  end
  if !set
    common_elms = (common_elms & crr_elms)
    set = common_elms.empty?
  end

  break if set && !pair
end

if pair
  puts "pairwise coprime"
elsif set
  puts "setwise coprime"
else
  puts "setwise coprime"
end