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 insert = table.insert

local n = read("n")

local a_max = 0
local a_t = {}
for i = 1, n do
	local a_i = read("n")
	a_t[i] = a_i
	a_max = max(a_i, a_max)
end


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

local is_setwise_coprime = false
local pre_gcd_num = a_t[1]
for i = 2, n do
	pre_gcd_num = gcd(max(pre_gcd_num, a_t[i]), min(pre_gcd_num, a_t[i]))
	if pre_gcd_num == 1 then
		is_setwise_coprime = true
	end
end


local dust_t = {}
for i = 2, a_max do
	if dust_t[i] == nil then
		local j = 2 * i
		while j <= a_max do
			if dust_t[j] == nil then
				dust_t[j] = i
			end
			j = j + i
		end
	end
end

local is_pairwise_coprime = true
local prime_factor_t = {}
for i = 1, n do
	local divided = a_t[i]
	local a_i_prime_t = {}
	while true do
		if prime_factor_t[divided] and divided ~= 1 then
			is_pairwise_coprime = false
			break
		elseif dust_t[divided] then
			insert(a_i_prime_t, divided)
			insert(a_i_prime_t, dust_t[divided])
			divided = divided // dust_t[divided]
		else
			insert(a_i_prime_t, divided)
			break
		end
	end
	for j = 1, #a_i_prime_t do
		prime_factor_t[a_i_prime_t[j]] = true
	end
	if not is_pairwise_coprime then
		break
	end
end

print(is_pairwise_coprime and "pairwise coprime" or (is_setwise_coprime and "setwise coprime" or "not coprime"))