CodeForces

Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex1. There is an integer number written on each vertex of the tree; the number written on vertexi is equal to a_i.

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 tox, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to0 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 a_i (1 ≤ i ≤ n,1 ≤ a_i ≤ 2·10⁵).

Each of next n - 1 lines contains two integer numbersx 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, wherei-th number equals to maximum possible beauty of vertexi.

Example

Input

26 21 2

Output

6 6 

Input

36 2 31 21 3

Output

6 6 6 

Input

110

Output

10 

分析:考虑长度为L的链,它的前缀gcd最多有log L个,然后我们要修改的那个点一定是两个不同的前缀gcd的交界,不然可以证明修改无意义,所以每次log n的复杂度枚举要修改的点就可以了。

#include <bits/stdc++.h>using namespace std;const int N = 2e5+5;int n,a[N],dep[N],pre[N],last[N],w[N][31],f[N][31];vector<int> G[N];void init(){for(int j = 1;(1<<j) <= n;j++) for(int i = 1;i <= n;i++) {f[i][j] = f[f[i][j-1]][j-1];w[i][j] = __gcd(w[f[i][j-1]][j-1],w[i][j-1]); }}int got(int x,int y){int tmp = a[x];for(int i = 20;i >= 0;i--) if(dep[f[x][i]] > dep[y])  { tmp = __gcd(tmp,w[x][i]); x = f[x][i]; }return tmp;}void dfs(int u,int fa){dep[u] = dep[fa] + 1;f[u][0] = fa;w[u][0] = fa ? __gcd(a[fa],a[u]) : a[u];if(!pre[fa]) pre[u] = a[u];else pre[u] = __gcd(pre[fa],a[u]);if(pre[u] != pre[fa]) last[u] = u;else last[u] = last[fa];for(int v : G[u]) if(v != fa) dfs(v,u);}int work(int u){int ans = -1;int p = last[u];while(p){int tmp = pre[f[p][0]];if(p != u){if(tmp) tmp = __gcd(tmp,got(u,p));else tmp = got(u,p);}ans = max(ans,tmp);p = last[f[p][0]];}return ans;}int main(){scanf("%d",&n);for(int i = 1;i <= n;i++) scanf("%d",&a[i]);for(int i = 1;i < n;i++){int u,v;scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);}dfs(1,0);init();cout<<a[1]<<" ";for(int i = 2;i <= n;i++) cout<<work(i)<<" ";}