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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int gcd(int a,int b){
	int i;
	int val;
	i=a;
	while(i>1){
		if(a%i==0 && b%i==0){
			val=i;
			return val;
		}
		i--;
	}
	val=1;
	return val;
}
int main(){
	int n;
	scanf("%d",&n);
	int a[n];
	for(int i=0;i<n;i++)
	scanf("%d",&a[i]);
	int pairCount=0;
	for(int i=0;i<n-1;i++){
		for(int j=i+1;j<n;j++){
			if(gcd(a[i],a[j])!=1){
				pairCount++;
			}
		}
	}
	if(pairCount==0){
		printf("pairwise coprime");
		return 0;
	}
	else{
		int temp=gcd(a[0],a[1]);
		int val;
		for(int i=2;i<n;i++){
			val=gcd(temp,a[i]);
		}
		if(val==1){
			printf("setwise coprime");
			return 0;
		}
		else{
			printf("not coprime");
			return 0;
		}
	}

}