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 (LuaJIT 2.1.0) (judge verdict: Accepted)

local function gcd(a,b)
	return b>0 and gcd(b,a%b) or a
end

----------

local n=io.read("n")
local a={}
for i=1,n do
	a[i]=io.read("n")
end

local t={1}
for i=2,1000000 do
	if not t[i] then
		for j=1,math.floor(1000000/i) do
			t[i*j]=i
		end
	end
end

local s={}
local pw=true
for i=1,n do
	local A=a[i]
	local T={}
	while A>1 do
		if not T[t[A]] then 
			T[t[A]]=true
		end
		A=math.floor(A/t[A])
	end
	for p in pairs(T) do
		if not s[p] then
			s[p]=true
		else
			pw=false
		end
	end
end

if pw then
	print("pairwise coprime")
else
	local sw=a[1]
	for i=2,n do
		sw=gcd(sw,a[i])
	end
	if sw==1 then
		print("setwise coprime")
	else
		print("not coprime")
	end
end