C. Ilya And The Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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.

ExamplesinputCopy

2
6 2
1 2

output

6 6 

inputCopy

3
6 2 3
1 2
1 3

output

6 6 6 

inputCopy

1
10

output

10 

題意：給一張圖，每個節點有一個權值，一個節點的beauty值=gcd(root  to now's weight)   【允許在一條路徑上有一個數為0】，求出每個節點的最大beauty值

思路：對於當前的數x，其beauty值有兩種情況

①  把這個數設為0，前面所有值是原值

②這個數是原來的數，前面的值都是原值 || 父親滿足的情況

#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;

const int N=2e5+6;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;
vector <int> edge[N];
set<int> st[N];
int a[N],vis[N];

void dfs(int u,int fa,int gcd){

    st[u].insert(__gcd(gcd,a[u])); // ALL GCD
    st[u].insert(gcd); // let a[u]=0                                                            
    for(set<int>::iterator it=st[fa].begin();it!=st[fa].end();it++){
        int t=*it;
        st[u].insert(__gcd(t,a[u]));
    }
    gcd=__gcd(gcd,a[u]);
    vis[u]=1;
    for(int i=0;i<edge[u].size();++i){
        int v=edge[u][i];
        if(!vis[v]){
            dfs(v,u,gcd);
        }
    }
}

int main(void){
    int n;
    cin >> n;
    for(int i=1;i<=n;i++)   scanf("%d",&a[i]);
    for(int i=1;i<=n-1;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    dfs(1,0,0);
    for(int i=1;i<=n;i++){
        if(i==1)    printf("%d",*st[i].rbegin());
        else printf(" %d",*st[i].rbegin());
    }
    return 0;
}