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：

求從根節點1到每一個節點x路徑上的gcd最大值，我們可以用一次機會使得路徑上的某個數變成0.

思路：

爆搜+set去重即可。因為每個數的因子數最多為logn個，那麼對應一條路徑上的gcd的值也不會多。

Ac程式碼：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
using namespace std;
vector<int>mp[250000];
int ans[250000];
int a[250000];

int gcd(int x,int y)
{
    return y==0?x:gcd(y,x%y);
}
void Dfs(int u,int from,set<int>now,int Val)
{
    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i];
        if(v==from)continue;
        set<int>nex;
        nex.insert(Val);
        set<int>::iterator it;
        for(it=now.begin();it!=now.end();it++)
        {
            nex.insert(gcd(*it,a[v]));
        }
        ans[v]=max(ans[v],*nex.rbegin());
        Dfs(v,u,nex,gcd(Val,a[v]));
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)mp[i].clear();
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=n-1;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            mp[x].push_back(y);
            mp[y].push_back(x);
        }
        ans[1]=a[1];
        set<int>temp;temp.clear();
        temp.insert(a[1]);
        temp.insert(0);
        Dfs(1,-1,temp,a[1]);
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
        printf("\n");
    }
}