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

public class Main
{
	static class Graph{
		LinkedList l[];
		int arr[],blue_in[],red_in[],V,red,blue;
		boolean rem[];

		public Graph(int v){
			V=v;
			l=new LinkedList[V+1];
			arr=new int[V+1];
			blue_in=new int[V+1];red_in=new int[V+1];
			//blue_out=new int[V+1];red_out=new int[V+1];
			blue=0;red=0;
			rem=new boolean[V+1];

			rem[0]=false;
			for(int i=1;i<=V;i++){
				l[i]=new LinkedList<Integer>();
				rem[i]=false;
			}
		}

		void addEdge(int u,int v){
			l[u].add(v);
			l[v].add(u);
		}

		void dfs(int u,int par){//1->red 2-> blue

			LinkedList<Integer> x=l[u];
			if(arr[u]==1)
				red_in[u]=1;
			else if(arr[u]==2)
				blue_in[u]=1;

			for(Integer child : x){
				if(child==par)
					continue;
				dfs(child,u);
				blue_in[u]+=blue_in[child];
				red_in[u]+=red_in[child];
			}
		}

		void fillArr()throws IOException{
			for(int i=1;i<=V;i++){
				if((arr[i]=ni())==1)
					red++;
				else if(arr[i]==2)
					blue++;
			}
		}

		long dfs1(int src){
			Queue<Integer> q=new LinkedList<>();
			q.add(src);
			long res=0l;
			int blue_out=0,red_out=0;
			boolean visited[]=new boolean[V+1];
			Arrays.fill(visited,false);
			
			while(!q.isEmpty()){
				int node=q.poll();
				visited[node]=true;
				LinkedList<Integer> x=l[node];
				for(Integer child: x){
					
					if(visited[child])
						continue;
					blue_out=Math.abs(blue-blue_in[child]);
					red_out=Math.abs(red-red_in[child]);

					q.add(child);

					if(blue_out!=0 && red_out!=0)
						continue;

					if(red_in[child]!=0 && blue_in[child]!=0)
						continue;
					//System.out.println(child+" "+node+" "+blue_out);
					res++;
				}
			}
			return res;
		}
	}
    public static void process()throws IOException
    {
    	int n=ni();
    	Graph g=new Graph(n);
    	g.fillArr();
    	for(int i=1;i<n;i++)
    		g.addEdge(ni(),ni());
    	
    	//System.out.println(g.blue);
    	g.dfs(1,0);
    	// System.out.println(Arrays.toString(g.red_in));
    	// System.out.println(Arrays.toString(g.blue_in));
    	pn(g.dfs1(1));
    }


   	static AnotherReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc=new AnotherReader();
        boolean oj = true;

    	oj = System.getProperty("ONLINE_JUDGE") != null;
    	if(!oj) sc=new AnotherReader(100);

        long s = System.currentTimeMillis();
        int t=1;
        while(t-->0)
            process();
        out.flush();
        if(!oj)
            System.out.println(System.currentTimeMillis()-s+"ms");
        System.out.close();  
    }

    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static void pni(Object o){out.println(o);System.out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    static boolean multipleTC=false;



/////////////////////////////////////////////////////////////////////////////////////////////////////////

    static class AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br=new BufferedReader(new InputStreamReader(System.in));}
    AnotherReader(int a)throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}
    
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}