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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;

import java.util.*;

public class A_GENERAL {
   
	// for fast output printing  : use printwriter or stringbuilder
	// remember to close pw using pw.close()
	static StringBuilder sb = new StringBuilder();
	static long seive_size = (long) 1e6;
	static String alpha = "abcdefghijklmnopqrstuvwxyz";
    static ArrayList<Integer> primes = new ArrayList<>();
    static boolean[] set = new boolean[(int) seive_size+1];
    static int n, m;
    static ArrayList<Integer>[] adj;
    static boolean[] visited;
    static ArrayDeque<Integer> q = new ArrayDeque<>();
    static final long MOD = 998244353;
    static long[][] arr;
    
    static long tot = 0;
    static long[] sub;
    static long[] a;
    static long ans = 0;
    
	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		MyScanner sc = new MyScanner();
		n = sc.nextInt();
		sub = new long[n+1];
		init(n);
		a = new long[n+1];
		for(int i = 1; i <= n; i++) {
			a[i] = sc.nextInt();
		}
		for(int i = 1; i <= n-1; i++) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			adj[u].add(v);
			adj[v].add(u);
		}
		build(1, -1, 0);
		dfs(1, -1, tot);
		out.println(ans);
		out.close();
	}
	public static long build(int u, int p, int d) {
		long res = a[u];
		tot += d*a[u];
		for(int v : adj[u]) {
			if(v != p) {
				res += build(v, u, d+1);
			}
		}
		return (sub[u] = res);
	}
	public static void dfs(int u, int p, long sum) {
		ans = Math.max(ans, sum);
		for(int v : adj[u]) {
			if(v != p) {
				dfs(v, u, sum-sub[v] + sub[1]-sub[v]);
			}
		}
	}
	public static void dfs(int u) {
		visited[u] = true;
		for(int v : adj[u]) {
			if(!visited[v]) {
				dfs(v);
			}
		}
	}
	
	public static class Pair {
		long f;
		long s;
		Pair(long f, long s) {
			this.f = f;
			this.s = s;
		}
	}
	public static void init(int n) {
    	adj = new ArrayList[n+1];
    	visited = new boolean[n+1];
    	for(int i = 1; i <= n; i++)
			adj[i] = new ArrayList();
    }
	// print string "s" multiple times 
	// prefer to use this function for multiple printing
	public static String mp(String s, int times) {
		return String.valueOf(new char[times]).replace("\0", s);
	}
	// take log with base 2
 	public static long log2(long k) {
 		return 63-Long.numberOfLeadingZeros(k);
 	}

 	// using lambda function for sorting
   	public static void lambdaSort() {
   		Arrays.sort(arr, (a, b) -> Double.compare(a[0], b[0]));
   	}
  	
  	// (n choose k) = (n/k) * ((n-1) choose (k-1))
  	public static long choose(long n, long k) {
 		return (k == 0) ? 1 : (n*choose(n-1, k-1))/k;
 	}

 	// just for keeping gcd function for other personal purposes
 	public static long gcd(long a, long b) {
 		return (a == 0) ? b : gcd(b%a, a);
 	}
 	
	public static long max(long... as) {
        long max = Long.MIN_VALUE;
        for (long a : as) max = Math.max(a, max);
        return max;
    }

    public static long min(int... as) {
        long min = Long.MAX_VALUE;
        for (long a : as) min = Math.min(a, min);
        return min;
    }
 	
    public static long modpow(long x, long n, long mod) {
    	if(n == 0) return 1%mod;
    	long u = modpow(x, n/2, mod);
    	u = (u*u)%mod;
    	if(n%2 == 1) u = (u*x)%mod;
    	return u;
    }
 	// ======================= binary search (lower and upper bound) =======================
	public static int lowerBound(long[] a, int x) {
		int lo = 0;
 		int hi = a.length-1;
 		int ans = -1;
 		while(lo <= hi) {
 			int mid = (lo+hi)/2;
 			if(x < a[mid]) {
 				hi = mid-1;
 			} else if(x > a[mid]) {
 				lo = mid+1;
 			} else if(lo != hi) {
 				 hi = mid-1; 		// for first occurrence
 				 ans = mid;
 			} else {
 				return mid;
 			}
 		}
 		return ans;
	}
	
	public static int upperBound(long[] a, long x) {
		int lo = 0;
		int hi = a.length-1;
		int ans = -1;
		while(lo <= hi) {
			int mid = (lo+hi)/2;
			if(x < a[mid]) {
				hi = mid-1;
			} else if(x > a[mid]) {
				lo = mid+1;
			} else if(lo != hi) {
				lo = mid+1;			// for last occurrence
				ans = mid;
			} else {
				return mid;
			}
		}
		return ans;
	}
	// ================================================================

	// ================== SEIVE OF ERATOSTHENES =======================
	// Complexity : O(N * log(log(N)))       ( almost O(N) )
	public static void generatePrimes() {
		// set.add(0);
		// set.add(1);
		Arrays.fill(set, true);
		set[0] = false;
		set[1] = false;
		for(int i = 2; i <= seive_size; i++) {
			if(set[i]) {
				for(long j = (long) i*i; j <= seive_size; j+=i) 
					set[(int)j] = false;
				primes.add(i);
			}
		}
	}
  
	public static boolean isPrime(long N) {
		if(N <= seive_size) return set[(int)N];
		for (int i = 0; i < (int)primes.size(); i++)
			if (N % primes.get(i) == 0) return false;
		return true;
	}
	// ===========================================================
  
 	// ================ Permutation of String ====================
 	public static void permute(String str) {
 		permute(str, 0, str.length()-1);
 	}
 	public static void permute(String str, int l, int r) 
    { 
        if (l == r) 
            System.out.println(str); 
        else
        { 
            for (int i = l; i <= r; i++) 
            { 
                str = swap(str,l,i); 
                permute(str, l+1, r); 
                str = swap(str,l,i); 
            } 
        } 
    }
	public static String swap(String a, int i, int j) 
    { 
        char temp; 
        char[] charArray = a.toCharArray(); 
        temp = charArray[i] ; 
        charArray[i] = charArray[j]; 
        charArray[j] = temp; 
        return String.valueOf(charArray); 
    } 
	// Union-find
	void makeSet(int parent[], int rank[]) {

		int i;
		for(i = 0; i < parent.length; i++) {
			parent[i] = i;
			rank[i] = 0;
		}
	}
	public static int find(int u, int parent[]) {

		if(parent[u] == u)	return u;

		int v = find(parent[u], parent);
		parent[u] = v;
		return v;

	}
	public static boolean connected(int u, int v, int[] parent) {
		return find(u, parent) == find(v, parent);
	}
	public static void Union(int u, int v, int rank[], int parent[]) {

		int x = find(u, parent);				//root of u
		int y = find(v, parent);				//root of v

		if(x == y)		return;
		if(rank[x] == rank[y]) {
			parent[y] = x;
			rank[x]++;
		}
		else if(rank[x] > rank[y]) {
			parent[y] = x;
		}
		else {
			parent[x] = y;
		}

	}
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			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;}
		int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

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

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

}