1092_F. Tree with Maximum Cost
You are given a tree consisting exactly of n vertices. Tree is a connected undirected graph with n-1 edges. Each vertex v of this tree has a value a_v assigned to it.

Let dist(x, y) be the distance between the vertices x and y. The distance between the vertices is the number of edges on the simple path between them.

Let's define the cost of the tree as the following value: firstly, let's fix some vertex of the tree. Let it be v. Then the cost of the tree is ∑_{i = 1}^{n} dist(i, v) ⋅ a_i.

Your task is to calculate the maximum possible cost of the tree if you can choose v arbitrarily.

Input

The first line contains one integer n, the number of vertices in the tree (1 ≤ n ≤ 2 ⋅ 10^5).

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the value of the vertex i.

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 one integer — the maximum possible cost of the tree if you can choose any vertex as v.

Examples

Input


8
9 4 1 7 10 1 6 5
1 2
2 3
1 4
1 5
5 6
5 7
5 8


Output


121


Input


1
1337


Output


0

Note

Picture corresponding to the first example: <image>

You can choose the vertex 3 as a root, then the answer will be 2 ⋅ 9 + 1 ⋅ 4 + 0 ⋅ 1 + 3 ⋅ 7 + 3 ⋅ 10 + 4 ⋅ 1 + 4 ⋅ 6 + 4 ⋅ 5 = 18 + 4 + 0 + 21 + 30 + 4 + 24 + 20 = 121.

In the second example tree consists only of one vertex so the answer is always 0.
```python
import sys
from collections import deque
from types import GeneratorType
sys.setrecursionlimit(200000)
input = sys.stdin.readline
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
 
    return wrappedfunc

n = int(input())
val = [int(i) for i in input().split()]
tree = [[] for i in range(n + 1)]
dp = [0 for i in range(n + 1)]
s = [0 for i in range(n + 1)]
ans = [0 for i in range(n + 1)]
for i in range(n - 1):
    a,b = map(int,input().split())
    tree[a].append(b)
    tree[b].append(a)
@bootstrap
def dfs1(node,dist,pd):

    for child in tree[node]:
        if child == pd:
            continue
        yield dfs1(child,dist + 1, node)
        dp[node] += dp[child]
        s[node] += s[child]
    dp[node] += val[node - 1] * dist
    s[node] += val[node - 1]
    yield dp[node]
dfs1(1,0,1)
q = deque(); ans[1] = dp[1]
for node in tree[1]:
    q.append((node,1))

while len(q) > 0:
    node,pd = q.popleft()
    sub_dp = ans[pd] - (dp[node] + s[node])
    added = s[1] - s[node]
    ans[node] = sub_dp + added + dp[node]
    for child in tree[node]:
        if child == pd:
            continue
        q.append((child,node))
print(max(ans))

```