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)

N = gets.to_i
A = gets.split.map(&:to_i)
Amax = A.max
D = Array.new(Amax + 1, &:itself)
(2 .. Integer.sqrt(Amax)).each do |x|
  next if D[x] < x
  s = x
  while s <= Amax
    D[s] = x
    s += x
  end
end

dp = Array.new(A.max + 1, 0)
A.each do |a|
  while D[a] > 1
    m = D[a]
    dp[m] += 1
    while D[a] == m
      a /= m
    end
  end
end

dpmax = dp.max
if dpmax <= 1
  puts "pairwise coprime"
elsif dpmax == N
  puts "non coprime"
else
  puts "setwise coprime"
end