#include <bits/stdc++.h>
using namespace std;
long long int n, ans;
vector<long long int> adj[200001];
long long int a[200001];
long long int sum_a[200001];
bool visited[200001];
void dfs(long long int s, long long int prof) {
  visited[s] = true;
  sum_a[s] = a[s];
  ans += (a[s] * prof);
  for (auto const &v : adj[s])
    if (!visited[v]) dfs(v, prof + 1), sum_a[s] += sum_a[v];
}
void dfs2(long long int s, long long int father, long long int at) {
  ans = max(ans, at);
  visited[s] = true;
  for (auto const &v : adj[s]) {
    if (!visited[v]) {
      long long int a1 = at, a2 = sum_a[v], a3 = sum_a[s];
      at -= sum_a[v], sum_a[s] -= sum_a[v], at += sum_a[s],
          sum_a[v] += sum_a[s];
      dfs2(v, s, at);
      at = a1, sum_a[v] = a2, sum_a[s] = a3;
    }
  }
}
signed main() {
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  if (n == 1) {
    cout << "0\n";
    return 0;
  }
  dfs(0, 0);
  memset(visited, 0, sizeof(visited));
  dfs2(0, 0, ans);
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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 int n, ans;
vector<long long int> adj[200001];
long long int a[200001];
long long int sum_a[200001];
bool visited[200001];
void dfs(long long int s, long long int prof) {
  visited[s] = true;
  sum_a[s] = a[s];
  ans += (a[s] * prof);
  for (auto const &v : adj[s])
    if (!visited[v]) dfs(v, prof + 1), sum_a[s] += sum_a[v];
}
void dfs2(long long int s, long long int father, long long int at) {
  ans = max(ans, at);
  visited[s] = true;
  for (auto const &v : adj[s]) {
    if (!visited[v]) {
      long long int a1 = at, a2 = sum_a[v], a3 = sum_a[s];
      at -= sum_a[v], sum_a[s] -= sum_a[v], at += sum_a[s],
          sum_a[v] += sum_a[s];
      dfs2(v, s, at);
      at = a1, sum_a[v] = a2, sum_a[s] = a3;
    }
  }
}
signed main() {
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  if (n == 1) {
    cout << "0\n";
    return 0;
  }
  dfs(0, 0);
  memset(visited, 0, sizeof(visited));
  dfs2(0, 0, ans);
  cout << ans << endl;
  return 0;
}

```