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 ProblemF_1 {

    public static InputStream inputStream = System.in;

    public static OutputStream outputStream = System.out;

    public static void main(String[] args) {
        MyScanner scanner = new MyScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        int n = scanner.nextInt();
        List<Integer> colors = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            colors.add(scanner.nextInt());
        }
        Graph graph = new Graph(n, colors);
        for (int i = 0; i < n - 1; i++) {
            graph.addEdge(scanner.nextInt(), scanner.nextInt());
        }

        graph.dfs();

        out.println(graph.countGoodEdges());




        out.flush();
    }

    private static class Graph {

        private Map<Integer, Node> nodes = new HashMap<>();

        private Graph(int n, List<Integer> colors) {
            for (int i = 1; i <= n; i++) {
                nodes.put(i, new Node(i, colors.get(i - 1)));
            }
        }

        private void addEdge(int x, int y) {
            nodes.get(x).children.add(nodes.get(y));
            nodes.get(y).children.add(nodes.get(x));
        }

        private void dfs() {
            nodes.get(1).dfs(nodes.get(1));
        }

        private int countGoodEdges() {
            return nodes.get(1).countGoodEdges(nodes.get(1), nodes.get(1).redsAmount, nodes.get(1).bluesAmount);
        }
    }

    private static class Node {

        private int label;

        private int color;

        private int redsAmount;

        private int bluesAmount;

        private List<Node> children = new ArrayList<>();

        private Node(int label, int color) {
            this.label = label;
            this.color = color;
        }

        private void dfs(Node parent) {
            for (Node child : children) {
                if (child.label != parent.label) {
                    child.dfs(this);
                    redsAmount += child.redsAmount;
                    bluesAmount += child.bluesAmount;
                }
            }
            if (color == 1) {
                redsAmount++;
            } else if (color == 2) {
                bluesAmount++;
            }
        }

        private int countGoodEdges(Node parent, int allRed, int allBlue) {
            int count = 0;
            for (Node child : children) {
                if (parent.label != child.label) {
                    count += child.countGoodEdges(this, allRed, allBlue);
                    if ((child.redsAmount == 0 || child.bluesAmount == 0) && (allRed == child.redsAmount || allBlue == child.bluesAmount)) {
                        count++;
                    }
                }
            }
            return count;
        }
    }

    private static class MyScanner {
        private BufferedReader bufferedReader;
        private StringTokenizer stringTokenizer;

        private MyScanner(InputStream inputStream) {
            bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        }

        private String next() {
            while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return stringTokenizer.nextToken();
        }

        private int nextInt() {
            return Integer.parseInt(next());
        }

        private long nextLong() {
            return Long.parseLong(next());
        }

        private double nextDouble() {
            return Double.parseDouble(next());
        }

        private String nextLine() {
            String str = "";
            try {
                str = bufferedReader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    private static class Pair<F, S> {

        private F first;

        private S second;

        public Pair() {}

        public Pair(F first, S second) {
            this.first = first;
            this.second = second;
        }
    }

    private static class Triple<F, S, T> {

        private F first;

        private S second;

        private T third;

        public Triple() {}

        public Triple(F first, S second, T third) {
            this.first = first;
            this.second = second;
            this.third = third;
        }
    }
}