#include <bits/stdc++.h>
using namespace std;
long long a, all[200010], subtreeSize[200010], _min, treeSum = 0;
vector<vector<long long> > vec(200010, vector<long long>());
long long calculateNodes(long long node, long long parent = -1,
                         long long _at = 0) {
  long long total = all[node], r = 0;
  for (long long i = 0; i < vec[node].size(); ++i) {
    long long ngb = vec[node][i];
    if (ngb == parent) continue;
    r += calculateNodes(ngb, node, _at + 1);
    total += subtreeSize[ngb];
  }
  subtreeSize[node] = total;
  return _at * all[node] + r;
}
void solveNode(long long node, long long parent, long long totalSum) {
  long long removeSum = subtreeSize[node];
  long long addSum = treeSum - subtreeSize[node];
  totalSum = totalSum - removeSum + addSum;
  _min = max(_min, totalSum);
  for (long long i = 0; i < vec[node].size(); ++i) {
    long long ngb = vec[node][i];
    if (ngb == parent) continue;
    solveNode(ngb, node, totalSum);
  }
}
int main() {
  cin >> a;
  for (long long i = 1; i <= a; ++i) {
    cin >> all[i];
    treeSum += all[i];
  }
  for (long long i = 0; i < a - 1; ++i) {
    long long temp1, temp2;
    cin >> temp1 >> temp2;
    vec[temp1].push_back(temp2);
    vec[temp2].push_back(temp1);
  }
  long long r = calculateNodes(1);
  _min = r;
  for (long long i = 0; i < vec[1].size(); ++i) solveNode(vec[1][i], 1, r);
  cout << _min << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 a, all[200010], subtreeSize[200010], _min, treeSum = 0;
vector<vector<long long> > vec(200010, vector<long long>());
long long calculateNodes(long long node, long long parent = -1,
                         long long _at = 0) {
  long long total = all[node], r = 0;
  for (long long i = 0; i < vec[node].size(); ++i) {
    long long ngb = vec[node][i];
    if (ngb == parent) continue;
    r += calculateNodes(ngb, node, _at + 1);
    total += subtreeSize[ngb];
  }
  subtreeSize[node] = total;
  return _at * all[node] + r;
}
void solveNode(long long node, long long parent, long long totalSum) {
  long long removeSum = subtreeSize[node];
  long long addSum = treeSum - subtreeSize[node];
  totalSum = totalSum - removeSum + addSum;
  _min = max(_min, totalSum);
  for (long long i = 0; i < vec[node].size(); ++i) {
    long long ngb = vec[node][i];
    if (ngb == parent) continue;
    solveNode(ngb, node, totalSum);
  }
}
int main() {
  cin >> a;
  for (long long i = 1; i <= a; ++i) {
    cin >> all[i];
    treeSum += all[i];
  }
  for (long long i = 0; i < a - 1; ++i) {
    long long temp1, temp2;
    cin >> temp1 >> temp2;
    vec[temp1].push_back(temp2);
    vec[temp2].push_back(temp1);
  }
  long long r = calculateNodes(1);
  _min = r;
  for (long long i = 0; i < vec[1].size(); ++i) solveNode(vec[1][i], 1, r);
  cout << _min << endl;
  return 0;
}

```