Problem: You are given a tree consisting of n vertices. A tree is a connected undirected graph with n-1 edges. Each vertex v of this tree has a color assigned to it (a_v = 1 if the vertex v is white and 0 if the vertex v is black).

You have to solve the following problem for each vertex v: what is the maximum difference between the number of white and the number of black vertices you can obtain if you choose some subtree of the given tree that contains the vertex v? The subtree of the tree is the connected subgraph of the given tree. More formally, if you choose the subtree that contains cnt_w white vertices and cnt_b black vertices, you have to maximize cnt_w - cnt_b.

Input

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

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the color of the i-th vertex.

Each of the next n-1 lines describes an edge of the tree. Edge i is denoted by two integers u_i and v_i, the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i).

It is guaranteed that the given edges form a tree.

Output

Print n integers res_1, res_2, ..., res_n, where res_i is the maximum possible difference between the number of white and black vertices in some subtree that contains the vertex i.

Examples

Input


9
0 1 1 1 0 0 0 0 1
1 2
1 3
3 4
3 5
2 6
4 7
6 8
5 9


Output


2 2 2 2 2 1 1 0 2 


Input


4
0 0 1 0
1 2
1 3
1 4


Output


0 -1 1 -1 

Note

The first example is shown below:

<image>

The black vertices have bold borders.

In the second example, the best subtree for vertices 2, 3 and 4 are vertices 2, 3 and 4 correspondingly. And the best subtree for the vertex 1 is the subtree consisting of vertices 1 and 3.

Solution (in JAVA):

Solution:
//package codeforces.round627div3;

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

public class MaximumWhiteSubtree {
    private static List<Integer>[] tree;
    private static Integer[] color;
    private static int[] subTreeMaxDiff;
    private static int[] res;
    public static void main(String[] args) {
//        try {
//            FastScanner in = new FastScanner(new FileInputStream("src/input.in"));
//            PrintWriter out = new PrintWriter(new FileOutputStream("src/output.out"));

        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

            solve(1, in, out);

//        } catch (IOException e) {
//            e.printStackTrace();
//        }
    }

    private static void solve(int q, FastScanner in, PrintWriter out) {
        for (int qq = 0; qq < q; qq++) {
            int n = in.nextInt();
            color = in.nextIntArray(n);
            Set<Integer>[] g = new Set[n];
            for(int i = 0; i < n; i++) {
                g[i] = new HashSet<>();
            }
            for(int i = 0; i < n - 1; i ++) {
                int u = in.nextInt() - 1;
                int v = in.nextInt() - 1;
                g[u].add(v);
                g[v].add(u);
            }
            //build tree using node 0 as root node
            tree = buildTree(g);
            subTreeMaxDiff = new int[n];
            res = new int[n];

            //first time dfs to compute each node's result when only considering each node as the root of each subtree and save results in subTreeMaxDiff
            //subTreeMaxDiff[0] is the final result for node 0 since we treat node 0 as the root of the entire tree
            firstTimeDfs(0);
            res[0] = subTreeMaxDiff[0];

            //second time dfs to compute the rest nodes' final results and save results in res.
            //for each node, update its children's results in the following way:
            //let's call v is a child node of parent node p, subTreeMaxDiff[v] has the all the contributions that are not from p.
            //compute p's contribution as res[p] - Math.max(subTreeMaxDiff[v], 0);
            //if p's contribution > 0, add it to subTreeMaxDiff[v] as the result for node v.
            secondTimeDfs(0);
            for(int i = 0; i < n; i++) {
                out.print(res[i] + " ");
            }
        }
        out.close();
    }

    private static List<Integer>[] buildTree(Set<Integer>[] g) {
        List<Integer>[] t = new List[g.length];
        for(int i = 0; i < t.length; i++) {
            t[i] = new ArrayList<>();
        }
        dfsTreeBuild(t, g, 0, -1);
        return t;
    }

    private static void dfsTreeBuild(List<Integer>[] tree, Set<Integer>[] g, int curr, int parent) {
        for(int to : g[curr]) {
            if(to == parent) {
                continue;
            }
            tree[curr].add(to);
            dfsTreeBuild(tree, g, to, curr);
        }
    }

    private static int firstTimeDfs(int idx) {
        int cnt = color[idx] == 1 ? 1 : -1;
        for(int v : tree[idx]) {
            int diff = firstTimeDfs(v);
            if(diff > 0) {
                cnt += diff;
            }
        }
        subTreeMaxDiff[idx] = cnt;
        return cnt;
    }

    private static void secondTimeDfs(int parent) {
        for(int v : tree[parent]) {
            int parentContribution = res[parent] - Math.max(subTreeMaxDiff[v], 0);
            res[v] = subTreeMaxDiff[v] + (parentContribution > 0 ? parentContribution : 0);
            secondTimeDfs(v);
        }
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

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

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }
    }
}