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.ArrayList;
import java.util.Scanner;

public class PSVM{
	// Find index of 0 to replaced with 1 to get maximum sequence
	// of continuous 1's
	public static int findIndexofZero(int[] A)
	{
		int max_count = 0;	// stores maximum number of 1's (including 0)
		int max_index = -1;   // stores index of 0 to be replaced

		int prev_zero_index = -1;   // stores index of previous zero
		int count = 0;  			// store current count of zeros

		// consider each index i of the array
		for (int i = 0; i < A.length; i++)
		{
			// if current element is 1
			if (A[i] == 1) {
				count++;
			}
			else
			// if current element is 0
			{
				// reset count to 1 + no. of ones to the left of current 0
				count = i - prev_zero_index;

				// update prev_zero_index to current index
				prev_zero_index = i;
			}

			// update maximum count & index of 0 to be replaced if required
			if (count > max_count)
			{
				max_count = count;
				max_index = prev_zero_index;
			}
		}

		// return index of 0 to be replaced or -1 if array contains all 1's
		return max_index;
	}
	
	public class Pair {
	    int red;
	    int blue;
	    Pair(int red, int blue) {
	        this.red = red;
	        this.blue = blue;
	    }
	}
	static int ans = 0;
	static int totBlue = 0;
	static int totRed = 0;
	Pair dfs(int u, int par, ArrayList< ArrayList<Integer> > grph, int[] col) {
	    
	    int r = (col[u] == 1) ? 1 : 0;
	    int b = (col[u] == 2) ? 1 : 0;
	    
	    for(int v : grph.get(u)) {
	        if(v != par) {
	            Pair cur = dfs(v, u, grph, col);
	            r += cur.red;
	            b += cur.blue;
	            ans += (cur.blue == totBlue && cur.red == 0) ? 1 : 0;
	            ans += (cur.blue == 0 && cur.red == totRed) ? 1 : 0;
	            
	        }
	    }
	    return new Pair(r, b);
	}

	// main function
	public static void main (String[] args)
	{
	   Scanner sc = new Scanner(System.in);
	   int n = sc.nextInt();
	   
	   int[] col = new int[n + 1];
	   for(int i = 1; i <= n; i++) {
	       col[i] = sc.nextInt();
	       if(col[i] == 1) {
	           totRed++;
	       } else if(col[i] == 2) {
	           totBlue++;
	       }
	   }
	   
	   ArrayList< ArrayList<Integer> > grph = new ArrayList< ArrayList<Integer> >();
	   
	   
	   for(int i = 0; i <= n; i++) {
	       grph.add(new ArrayList<Integer>());
	   }
	   
	   for(int i = 0; i < n - 1; i++) {
	       int u = sc.nextInt();
	       int v = sc.nextInt();
	       ArrayList<Integer> curU =  grph.get(u);
	       curU.add(v);
	       grph.set(u, curU);
           ArrayList<Integer> curV =  grph.get(v);
           curV.add(u);
	       grph.set(v, curV);
	       
	   }
	   PSVM car = new PSVM();
	   car.dfs(1, 0, grph, col);
	   System.out.println(ans);
	}
}