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 Lua (Lua 5.3.5) (judge verdict: Accepted)

local read = io.read
local max = math.max
local min = math.min

local max_n = 1000003


local function gcd(x, y)
	if y == 0 then
		return x
	else
		return gcd(y, x % y)
	end
end

local function main()
	local n = read("n")
	local num_t = {}
	for i = 1, max_n do
		num_t[i] = 0
	end
	local z = 0
	for i = 1, n do
		local x = read("n")
		num_t[x] = num_t[x] + 1
		z = gcd(max(z, x), min(z, x))
	end
	if z ~= 1 then
		print("not coprime")
		return
	end

	for i = 2, max_n do
		local cnt = 0
		for j = i, max_n, i do
			cnt = cnt + num_t[j]
		end
		if cnt > 1 then
			print("setwise coprime")
			return
		end
	end
	print("pairwise coprime")
end

main()