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 Solution {
    static ArrayList<Integer> adjList[];
    static int redCount = 0;
    static int blueCount = 0;
    static int n;
    static int res = 0;
    static boolean[] visited;
    static int[] color;
    static Pair [] pairs;

//    static void dfs(int node, int red, int blue) {
//        visited[node] = true;
//        for (Integer child : adjList[node]) {
//            if (!visited[child]) {
//                if (color[child] == 1)
//                    dfs(child, red + 1, blue);
//                else if (color[child] == 2)
//                    dfs(child, red, blue + 1);
//                else dfs(child, red, blue);
//            }
//        }
//
//        if (red == redCount && blue == 0) res++;
//        if (red == 0 && blue == blueCount) res++;
//    }

    static Pair dfs(int node) {
        visited[node] = true;
        int red = 0;
        int blue = 0;
        for (Integer child : adjList[node]) {
            if (!visited[child]) {
                Pair p = dfs(child);
                red += p.x;
                blue += p.y;
            }
        }
        if (color[node] == 1) {
            red++;
        }
        if (color[node] == 2) {
            blue++;
        }
        // I lose the red and blue once the backtrack finish
        // Important: need to save the state of the subtree for later use

         return  pairs[node] = new Pair(red, blue);
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);

        n = sc.nextInt();
        adjList = new ArrayList[n + 1];
        pairs = new Pair[n + 1];

        color = new int[n + 1];
        visited = new boolean[n + 1];
        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
            int num = sc.nextInt();
            color[i] = num;
            if (num == 1) redCount++;
            else if (num == 2) blueCount++;
        }
        adjList[n] = new ArrayList<>();

        for (int i = 0; i < n - 1; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            adjList[a].add(b);
            adjList[b].add(a);
        }
        adjList[n].add(0);


        dfs(n);

        for (int i = 0; i < n; i++) {
            if (pairs[i].x * pairs[i].y == 0 && (redCount - pairs[i].x) * (blueCount - pairs[i].y) == 0) {
                res++;
            }
        }
        System.out.println(res);
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream System) throws FileNotFoundException {
            br = new BufferedReader(new InputStreamReader(System));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

        public void waitForInput() {
            for (long i = 0; i < 3e9; i++)
                ;
        }
    }
    static class Pair {
        int x;
        int y;

        Pair(int start, int end) {
            this.x = start;
            this.y = end;
        }
    }

}