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:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300IQ as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1092F
   {
      static int[] arr;
      static ArrayDeque<Integer>[] edges;
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         arr = new int[N+1];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr[i+1] = Integer.parseInt(st.nextToken());
         edges = new ArrayDeque[N+1];
         for(int i=1; i <= N; i++)
            edges[i] = new ArrayDeque<Integer>();
         for(int i=0; i < N-1; i++)
         {
             st = new StringTokenizer(infile.readLine());
             int a = Integer.parseInt(st.nextToken());
             int b = Integer.parseInt(st.nextToken());
             edges[a].add(b); edges[b].add(a);
         }
         total = 0L;
         for(int x: arr)
            total += x;
         int[] dist = new int[N+1];
         long[] sum = new long[N+1];
         dfs(1, 0, sum, dist);
         long curr = 0L;
         for(int i=1; i <= N; i++)
            curr += (long)arr[i]*dist[i];
         res = curr;
         reroot(1, 0, curr, sum);
         System.out.println(res);
      }
      static long res;
      static long total;
      public static void reroot(int curr, int par, long val, long[] sum)
      {
         for(int next: edges[curr])
            if(next != par)
            {
               long temp = val+total-2*sum[next];
               res = Math.max(res, temp);
               reroot(next, curr, temp, sum);
            }
      }
      public static void dfs(int curr, int par, long[] sum, int[] dist)
      {
         long res = arr[curr];
         for(int next: edges[curr])
            if(next != par)
            {
               dist[next] = dist[curr]+1;
               dfs(next, curr, sum, dist);
               res += sum[next];
            }
         sum[curr] = res;
      }
   }