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 PYTHON3):

Solution:
import sys
input = sys.stdin.readline

N = int(input())
Color = list(map(int, input().split()))
Adj = [[] for _ in range(N)]

for _ in range(N-1):
    a, b = map(lambda x: int(x)-1, input().split())
    Adj[a].append(b)
    Adj[b].append(a)

Parent = [0]*N
Parent[0] = -1
rank = [0]*N

Q = [0]
visited = {0}
depth = 1

# BFS
while Q:
    P = []  # next iteration
    for i in Q:
        for j in Adj[i]:
            if j in visited:
                continue
            rank[j] = depth  # depth
            Parent[j] = i  # parent
            visited.add(j)
            P.append(j)

    Q = P
    depth += 1

# set of [depth,node]
rankk = [[rank[i], i] for i in range(N)]
rankk.sort()

# above code calc depth of each nodes from Parent(0)
Ans = [0]*N  # ans

# start from farthest nodes(leave)
for i in reversed(range(1, N)):
    k = rankk[i][1]  # id
    if Color[k] == 0:  # node k is black
        Ans[k] -= 1  # minus 1 as itself blakc cnt
    else:
        Ans[k] += 1  # node k is white
    # if children value (Ans[k]) <0 , not use children ,just cut off children
    Ans[Parent[k]] += max(0, Ans[k])

if Color[0] == 0:
    Ans[0] -= 1
else:
    Ans[0] += 1
# above code
# calc max diff on each nodes when 0 is Parent and consider childrens of the node.

# change Parent and calc value of each node from top(Parent 0)
for i in range(1, N):
    k = rankk[i][1]
    Ans[k] = max(Ans[k], min(Ans[k], 0)+Ans[Parent[k]])
    # compare two setting
    # 1. Ans[k]: ignore parent value
    # 2. min(Ans[k],0) + Ans[Parent[k]]: use parent

print(*Ans)