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.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.Stack;

public class ROUGH{
	static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
  
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static int mod = (int) (1e9+7),n;
	static long cf = 998244353;
    static long MAX = (long)(2e5),m,leaves;
    public static List<Integer>[] edges,prime;
    public static Map<Integer,Integer>[] dp;
    public static int[] a;
  
	public static void main(String[] args) throws IOException {
		Reader sc = new Reader();
		int n = sc.nextInt();
		edges = new ArrayList[n+1];
		prime = new ArrayList[n+1]; 
		dp = new HashMap[n+1];
		a = new int[n+1];
		for(int i=0;i<=n;++i) {
			edges[i] = new ArrayList<>();
			prime[i] = new ArrayList<>(); 
			dp[i] = new HashMap<>(); // key-> prime fac, val-> max path length
		}
		for(int i=1;i<=n;++i) {
			a[i] = sc.nextInt();
			int x = a[i];
			for(int j=2;j*j<=MAX;++j) {
				if(x%j == 0) {
					prime[i].add(j);
					while(x%j == 0) x/=j;
				}
			}
			if(x > 1) prime[i].add(x);
			Collections.sort(prime[i]);
		}
		
		for(int i=2;i<=n;++i) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			edges[u].add(v);
			edges[v].add(u);
		}
		len = 0;
		dfs(1,0);
		out.print(len);
	    out.close();
	}
    
	static int len;
	private static void dfs(int v, int par) {
		for(int p : prime[v]) dp[v].put(p,1);
		
		for(int child : edges[v]) {
			if(child != par) {
				dfs(child,v);
				for(int p : prime[v]) {
					int ui = dp[v].get(p); // initially 1 but it contain max1 child len
					int vi = dp[child].getOrDefault(p,0);
					len = Math.max(len, ui + vi);
					ui = Math.max(ui, vi + 1);
					dp[v].put(p,ui);
				}
			}
		}
		for(int p : prime[v]) len = Math.max(len,dp[v].get(p));
	}
}