C. Ilya And The Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex i is equal to ai.

Ilya believes that the beauty of the vertex x is the greatest common divisor of all numbers written on the vertices on the path from the root to x

, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.

For each vertex the answer must be considered independently.

The beauty of the root equals to number written on it.

Input

First line contains one integer number n — the number of vertices in tree (1 ≤ n ≤ 2·105).

Next line contains n integer numbers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 2·105).

Each of next n - 1 lines contains two integer numbers x and y (1 ≤ x, y ≤ n, x ≠ y), which means that there is an edge (x, y) in the tree.

Output

Output n numbers separated by spaces, where i-th number equals to maximum possible beauty of vertex i.

Examples input

2
6 2
1 2

output

6 6 

input

3
6 2 3
1 2
1 3

output

6 6 6 

input

1
10

output

10 

題目大意:有一個樹，樹上的每個點都有一個權值，對於一個點的魅力值來說，是他到根點1的路徑上的所有點權值的最大公約數，對於每個路徑上的點，你可以選擇唯一點將他變為0或者不做改變，現在希望你讓每個點都獲得儘可能大的魅力值，輸出每個點的魅力值

解題思路:這道題可以在dfs遍歷的時候，若是某個權值被修改，一定是這個權值的因子過小，導致這個路徑上的魅力值變小，那麼我們就列舉當前點的因子，判斷對於他的因子來說，是否在他之前的路徑中出現次數大於等於深度-2，若是超過，則說明這個因子肯定會存在，那麼就要這個因子，與x之前所有點gcd的值取max

深度-2的原因是，它意味著這個因子出現的次數僅僅在之前路徑上的某一個權值中不存在，若是這個因子很大，那麼我們就可以將這個權值變為0

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;

vector<int> tu[200005];
int b[200005],a[200005],check[200005],cnt[200005];
int n;
int gcd(int x, int y){
	if(y == 0) return x;
	else return gcd(y, x % y);
}
void dfs(int son,int fa,int deep)
{
	int i,flag;
	b[son]=gcd(a[son],b[fa]);
	check[son]=b[fa];
	for(i=1;i*i<=a[son];i++)
	{
		if(a[son]%i==0)
		{
			flag=i;
			if(cnt[flag]>=deep-2) check[son]=max(flag,check[son]);
			flag=a[son]/i;
			if(cnt[flag]>=deep-2) check[son]=max(flag,check[son]);
			if(i*i==a[son])
				cnt[i]++;
			else
				cnt[i]++,cnt[a[son]/i]++;
		}
	}
	for(i=0;i<tu[son].size();i++)
	{
		int k=tu[son][i];
		if(k==fa)
			continue;
		dfs(k,son,deep+1);
	}
	for(i=1;i*i<=a[son];i++)
	{
		if(a[son]%i==0)
		{
			if(i*i==a[son])
				cnt[i]--;
			else
				cnt[i]--,cnt[a[son]/i]--;
		}
	}
}
int main()
{
	int x,y,i;
	cin>>n;
	for(i=1;i<=n;i++) cin>>a[i];
	for(i=1;i<n;i++)
	{
		cin>>x>>y;
		tu[x].push_back(y);
		tu[y].push_back(x);
	}
	dfs(1,0,1);
	for(i=1;i<=n;i++) cout<<check[i]<<" ";
	cout<<endl;
}