You are given a tree consisting of n vertices. A number is written on each vertex; the number on vertex i is equal to a_i.

Let's denote the function g(x, y) as the greatest common divisor of the numbers written on the vertices belonging to the simple path from vertex x to vertex y (including these two vertices). Also let's denote dist(x, y) as the number of vertices on the simple path between vertices x and y, including the endpoints. dist(x, x) = 1 for every vertex x.

Your task is calculate the maximum value of dist(x, y) among such pairs of vertices that g(x, y) > 1.

Input

The first line contains one integer n — the number of vertices (1 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the numbers written on vertices.

Then n - 1 lines follow, each containing two integers x and y (1 ≤ x, y ≤ n, x ≠ y) denoting an edge connecting vertex x with vertex y. It is guaranteed that these edges form a tree.

Output

If there is no pair of vertices x, y such that g(x, y) > 1, print 0. Otherwise print the maximum value of dist(x, y) among such pairs.

Examples

Input

3
2 3 4
1 2
2 3


Output

1


Input

3
2 3 4
1 3
2 3


Output

2


Input

3
1 1 1
1 2
2 3


Output

0
import java.util.*;
import java.io.*;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    int[] spf, c;
    void pre() throws Exception{
        spf = new int[MX];
        for(int i = 2; i< MX; i++)
            if(spf[i]==0)
                for(int j = i; j< MX; j+=i)
                    if(spf[j]==0)
                        spf[j] = i;
    }
    void solve(int TC) throws Exception{
        int n = ni();
        int[] a = new int[n];
        for(int i = 0; i< n; i++)a[i] = ni();
        int[][] e = new int[n-1][];
        for(int i = 0; i< n-1; i++)e[i] = new int[]{ni()-1, ni()-1};
        HashMap<Integer, Integer>[] map = new HashMap[n];
        for(int i = 0; i< n; i++)map[i] = new HashMap<>();
        int[][] g = makeU(n,e);
        pn(dfs(g, map,a, 0, -1));
    }
    int dfs(int[][] g, HashMap<Integer, Integer>[] map, int[] a, int u, int p){
        int ans = 0;
        for(int v:g[u])if(v!=p)ans = Math.max(ans, dfs(g, map, a, v, u));
        int x = a[u];
        while(x>1){
            int pr = spf[x];
            while(x%pr==0)x/=pr;
            int mx1 = 0, mx2 = 0;
            for(int v:g[u]){
                if(a[v]%pr==0){
                    int c = map[v].getOrDefault(pr, 0);
                    if(c>mx1){
                        mx2 = mx1;
                        mx1 = c;
                    }else if(c>mx2)mx2 = c;
                }
            }
            ans = Math.max(ans, mx1+mx2+1);
            map[u].put(pr, mx1+1);
        }
        return ans;
    }
    int[][] makeU(int n, int[][] edge){
        int[][] g = new int[n][];int[] cnt = new int[n];
        for(int i = 0; i< edge.length; i++){cnt[edge[i][0]]++;cnt[edge[i][1]]++;}
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< edge.length; i++){
            g[edge[i][0]][--cnt[edge[i][0]]] = edge[i][1];
            g[edge[i][1]][--cnt[edge[i][1]]] = edge[i][0];
        }
        return g;
    }
    //SOLUTION END
    void hold()throws Exception{throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)4e18;
    final int INF = (int)2e9, MX = (int)1e6+1;
    DecimalFormat df = new DecimalFormat("0.000000000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        pre();for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }  
            return str;
        }
    }
}    