#include <bits/stdc++.h>
using namespace std;
long long int sum[200007], res = 0, ans = INT_MIN;
void dfs(vector<vector<int>> &list, int index, int parent, long long int arr[],
         int a, int h) {
  if (a) {
    res += (h * arr[index]);
    sum[index] = arr[index];
    for (int i = 0; i < list[index].size(); i++) {
      int b = list[index][i];
      if (b != parent) {
        dfs(list, b, index, arr, a, h + 1);
        sum[index] += sum[b];
      }
    }
  } else {
    ans = max(ans, res);
    for (int i = 0; i < list[index].size(); i++) {
      int b = list[index][i];
      if (b != parent) {
        sum[index] -= sum[b];
        res -= sum[b];
        sum[b] += sum[index];
        res += sum[index];
        dfs(list, b, index, arr, a, h);
        res -= sum[index];
        sum[b] -= sum[index];
        res += sum[b];
        sum[index] += sum[b];
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long int arr[n + 1];
  for (int i = 0; i < n; i++) cin >> arr[i + 1];
  vector<vector<int>> list(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    list[a].push_back(b);
    list[b].push_back(a);
  }
  dfs(list, 1, -1, arr, 1, 0);
  dfs(list, 1, -1, arr, 0, 1);
  cout << ans;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
long long int sum[200007], res = 0, ans = INT_MIN;
void dfs(vector<vector<int>> &list, int index, int parent, long long int arr[],
         int a, int h) {
  if (a) {
    res += (h * arr[index]);
    sum[index] = arr[index];
    for (int i = 0; i < list[index].size(); i++) {
      int b = list[index][i];
      if (b != parent) {
        dfs(list, b, index, arr, a, h + 1);
        sum[index] += sum[b];
      }
    }
  } else {
    ans = max(ans, res);
    for (int i = 0; i < list[index].size(); i++) {
      int b = list[index][i];
      if (b != parent) {
        sum[index] -= sum[b];
        res -= sum[b];
        sum[b] += sum[index];
        res += sum[index];
        dfs(list, b, index, arr, a, h);
        res -= sum[index];
        sum[b] -= sum[index];
        res += sum[b];
        sum[index] += sum[b];
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long int arr[n + 1];
  for (int i = 0; i < n; i++) cin >> arr[i + 1];
  vector<vector<int>> list(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    list[a].push_back(b);
    list[b].push_back(a);
  }
  dfs(list, 1, -1, arr, 1, 0);
  dfs(list, 1, -1, arr, 0, 1);
  cout << ans;
}

```