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

public class Main {
    static final int N = 200005;
    static List<Integer> graph[] = new List[N];
    static int n;
    static int[] a = new int[N], dw = new int[N], up = new int[N];
    public static void main(String[] args) throws IOException {
        MScanner in = new MScanner(System.in);
        n = in.nextInt();
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
            if (a[i] == 0)
                a[i] = -1;
            graph[i] = new ArrayList<Integer>();
        }
        for (int i = 1; i < n; i++) {
            int u = in.nextInt(), v = in.nextInt();
            graph[u].add(v);
            graph[v].add(u);
        }
        dfs_dw(1, 1);
        dfs_up(1, 1);
        for (int i = 1; i <= n; i++) {
            System.out.print(dw[i] + up[i] + " ");
        }
    }
    static void dfs_dw(int v, int pr) {
        dw[v] = a[v];
        for (int to : graph[v]) {
            if (to == pr)
                continue;
            dfs_dw(to, v);
            dw[v] += Math.max(dw[to], 0);
        }
    }
    static void dfs_up(int v, int pr) {
        int ret = up[v] + dw[v];
        for (int to : graph[v]) {
            if (to == pr)
                continue;
            up[to] = Math.max(0, ret - Math.max(dw[to], 0));
            dfs_up(to, v);
        }
    }
    static class MScanner {
        StringTokenizer st;
        BufferedReader br;
        public MScanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public MScanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public int[] takearr(int n) throws IOException {
            int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
            return in;
        }
        public long[] takearrl(int n) throws IOException {
            long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
            return in;
        }
        public Integer[] takearrobj(int n) throws IOException {
            Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
            return in;
        }
        public Long[] takearrlobj(int n) throws IOException {
            Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
            return in;
        }
        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() throws InterruptedException {
            Thread.sleep(3000);
        }
    }
}