#include <bits/stdc++.h>
using namespace std;
vector<int> Graph[200005];
int Array[200005];
long long dp[200005], Cost[200005];
void dfs(int root, int parent, int dist) {
  dp[root] = Array[root];
  Cost[1] += dist * 1LL * Array[root];
  for (int i = 0; i < Graph[root].size(); ++i) {
    if (Graph[root][i] != parent) {
      dfs(Graph[root][i], root, dist + 1);
      dp[root] += dp[Graph[root][i]];
    }
  }
}
void Find_Ans(int root, int parent) {
  Cost[root] = Cost[parent] + dp[1] - 2 * dp[root];
  for (int i = 0; i < Graph[root].size(); ++i) {
    if (Graph[root][i] != parent) Find_Ans(Graph[root][i], root);
  }
}
int main() {
  int N, i, u, v;
  scanf("%d", &N);
  for (i = 1; i <= N; ++i) scanf("%d", Array + i);
  for (i = 1; i < N; ++i) {
    scanf("%d%d", &u, &v);
    Graph[u].emplace_back(v);
    Graph[v].emplace_back(u);
  }
  dfs(1, 0, 0);
  for (i = 0; i < Graph[1].size(); ++i) Find_Ans(Graph[1][i], 1);
  long long Ans = 0;
  for (i = 1; i <= N; ++i) Ans = max(Ans, Cost[i]);
  printf("%lld\n", Ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> Graph[200005];
int Array[200005];
long long dp[200005], Cost[200005];
void dfs(int root, int parent, int dist) {
  dp[root] = Array[root];
  Cost[1] += dist * 1LL * Array[root];
  for (int i = 0; i < Graph[root].size(); ++i) {
    if (Graph[root][i] != parent) {
      dfs(Graph[root][i], root, dist + 1);
      dp[root] += dp[Graph[root][i]];
    }
  }
}
void Find_Ans(int root, int parent) {
  Cost[root] = Cost[parent] + dp[1] - 2 * dp[root];
  for (int i = 0; i < Graph[root].size(); ++i) {
    if (Graph[root][i] != parent) Find_Ans(Graph[root][i], root);
  }
}
int main() {
  int N, i, u, v;
  scanf("%d", &N);
  for (i = 1; i <= N; ++i) scanf("%d", Array + i);
  for (i = 1; i < N; ++i) {
    scanf("%d%d", &u, &v);
    Graph[u].emplace_back(v);
    Graph[v].emplace_back(u);
  }
  dfs(1, 0, 0);
  for (i = 0; i < Graph[1].size(); ++i) Find_Ans(Graph[1][i], 1);
  long long Ans = 0;
  for (i = 1; i <= N; ++i) Ans = max(Ans, Cost[i]);
  printf("%lld\n", Ans);
  return 0;
}

```