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:

public class Main {


  private static void solve() {
    int n = ni();
    long[] a = nal(n);
    int[] from = new int[n - 1];
    int[] to = new int[n - 1];
    for (int i = 0; i < n - 1; i++) {
      from[i] = ni() - 1;
      to[i] = ni() - 1;
    }
    int[][] g = packU(n, from, to);

    long[][] dp = new long[n][2];
    dfs(0, -1, a, g, dp);
    long ret = dfs2(0, -1, new long[2], a, g, dp);

    System.out.println(ret);
  }



  private static long dfs2(int now, int pre, long[] dpar, long[] a, int[][] g, long[][] dp) {
    long ret = dp[now][1] + dpar[1];

    
    for (int next : g[now]) {
      if (next == pre)
        continue;

      long[] ndp = new long[2];
      ndp[0] = dpar[0] + dp[now][0] - dp[next][0];
      ndp[1] = (dp[now][1] + dpar[1]) - (dp[next][1] + dp[next][0]) + ndp[0];
      long v = dfs2(next, now, ndp, a, g, dp);
      ret = Math.max(ret, v);
    }
    return ret;

  }



  private static void dfs(int now, int pre, long[] a, int[][] g, long[][] dp) {
    for (int next : g[now]) {
      if (next == pre)
        continue;

      dfs(next, now, a, g, dp);
      dp[now][0] += dp[next][0];
      dp[now][1] += dp[next][1];
    }
    dp[now][1] += dp[now][0];
    dp[now][0] += a[now];
  }



  public static int[][] packU(int n, int[] from, int[] to) {
    return packU(n, from, to, from.length);
  }

  public static int[][] packU(int n, int[] from, int[] to, int sup) {
    int[][] g = new int[n][];
    int[] p = new int[n];
    for (int i = 0; i < sup; i++)
      p[from[i]]++;
    for (int i = 0; i < sup; i++)
      p[to[i]]++;
    for (int i = 0; i < n; i++)
      g[i] = new int[p[i]];
    for (int i = 0; i < sup; i++) {
      g[from[i]][--p[from[i]]] = to[i];
      g[to[i]][--p[to[i]]] = from[i];
    }
    return g;
  }

  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }

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

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

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

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

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}