#include<iostream>
using namespace std;
int n,a[101],num;
int lcm(int a,int b){//求最小公倍数
	int r=a%b,x=a,y=b;
	while(r){
		a=b;
		b=r;
		r=a%b;
	}
	return x*y/b;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	int num=a[1];
	for(int i=2;i<=n;i++)num=lcm(num,a[i]);//求所有数的最小公倍数
	for(int i=1;i<=n;i++)cout<<num/a[i]<<endl;//输出
	return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[N];
int main()
{
	int	n;
	cin >> n;
	int sum = 1;
	for(int i = 1 , x , y ; i <= n ; i++)
	{
		cin >> a[i];
		y = sum;
		x = a[i];
		sum *= x;
		while(y)
		{
			int t = x%y;
			x = y;
			y = t;
		}
		sum /= x;
	}
	for(int i = 1 ; i <= n ; i++)
	{
		cout << sum / a[i] << endl;
	}
}