Problem: You are given an undirected tree of n vertices. 

Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.

You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.

How many nice edges are there in the given tree?

Input

The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.

The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.

Output

Print a single integer — the number of nice edges in the given tree.

Examples

Input


5
2 0 0 1 2
1 2
2 3
2 4
2 5


Output


1


Input


5
1 0 0 0 2
1 2
2 3
3 4
4 5


Output


4


Input


3
1 1 2
2 3
1 3


Output


0

Note

Here is the tree from the first example:

<image>

The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.

Here is the tree from the second example:

<image>

Every edge is nice in it.

Here is the tree from the third example:

<image>

Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;







public class Main {
	static final int N=300005;
	static final int M=200005;
	

    static class FastScanner{
           BufferedReader br;
           StringTokenizer st;
           public FastScanner(InputStream in) {
        	   br=new BufferedReader(new InputStreamReader(in),16384);
        	   eats("");
           }
           public void eats(String s) {
        	   st=new StringTokenizer(s);
           }
           public String nextLine() {
        	   try {
				return br.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				return null;
			}
           }
           public boolean hasNext() {
        	   while(!st.hasMoreElements()) {
        		   String s =nextLine();
        		   if(s==null)return false;
        		   eats(s);
        	   }
        	   return true;
           }
           public String next() {
        	   hasNext();
        	   return st.nextToken();
           }
           public int nextInt() {
        	   return Integer.parseInt(next());
           }
    }
    
    static  FastScanner cin =new FastScanner(System.in);
    static  PrintWriter out =new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static class node{
    	int v;
    	int nxt;
    };
    static node Edge[]=new node [2*N];
    static int val[]=new int [N];
    static int head[]=new int[2*N];
    static int cnt=0;
    static void add(int x,int y) {
    	   Edge[cnt].v=y;
    	   Edge[cnt].nxt=head[x];
    	   head[x]=cnt++;
    	   return ;
    }
    static int du[]=new int [N];
    static int dp[]=new int [N];
    static int dp1[]=new int [N]; 
    static int ans=0,cnt1=0,cnt2=0;
    static int vis[]=new int [N];
    static void dfs(int now,int fa) {
    	    
    	    int c=0;
            for(int i=head[now];i!=-1;i=Edge[i].nxt) {
            	int v=Edge[i].v;
            	if(v==fa)continue;
            	c++;
            	dfs(v,now);
            	dp[now]+=dp[v];
            	dp1[now]+=dp1[v];
            }
            if(c==0) {
            	if(val[now]==1) {
            		dp[now]=1;
            	}
            	else if(val[now]==2) {
            		dp1[now]=1;
            	}
            }
           // out.println(now+" "+dp[now]+" "+dp1[now]);
           
    }
    
	public static void main(String[] args) {
		 int n;
		 n=cin.nextInt();
		 
		 for(int i=0;i<=2*n;i++) {
			 Edge[i]=new node();
			 head[i]=-1;
		 }
		 for(int i=1;i<=n;i++) {
			 val[i]=cin.nextInt();
		     
			 if(val[i]==1) {
				 cnt1++;
				 dp[i]=1;
			 }
			 else if(val[i]==2){
				 dp1[i]=1;
				 cnt2++;
			 }
		 }
		 int x,y;
		 for(int i=1;i<n;i++) {
			 x=cin.nextInt();
			 y=cin.nextInt();
			 add(x, y);
			 add(y, x);
			 
		 }
		
		 dfs(1,-1);
		 for(int i=1;i<=n;i++) {
			//out.println(dp[i]+" "+dp1[i]);
			 if((dp[i]==cnt1&&dp1[i]==0)||(dp[i]==0&&dp1[i]==cnt2)) {
				 ans++;
			 }
		 }
    	 out.println(ans);
		 out.flush();
	     return ;

	}

}