#include <bits/stdc++.h>
using namespace std;
long long rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long mod = 1e9 + 7;
const long long N = 300050;
long long n, a[N], dis[N], val[N], nodeans[N], ans = 0;
vector<vector<long long> > adj(N);
vector<long long> vis(N);
void dfs(long long v) {
  vis[v] = 1;
  val[v] += a[v];
  for (long long u : adj[v]) {
    if (!vis[u]) {
      dis[u] = dis[v] + 1;
      dfs(u);
      val[v] += val[u];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long test = 1;
  while (test--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1);
    for (long long i = 1; i <= n; i++) {
      ans += dis[i] * a[i];
      vis[i] = 0;
    }
    nodeans[1] = ans;
    long long tot = val[1];
    queue<long long> q;
    q.push(1);
    while (q.size()) {
      long long v = q.front();
      vis[v] = 1;
      q.pop();
      for (long long u : adj[v]) {
        if (!vis[u]) {
          nodeans[u] = nodeans[v] + (tot - 2 * val[u]);
          ans = max(ans, nodeans[u]);
          q.push(u);
        }
      }
    }
    cout << ans << '\n';
  }
  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 rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long mod = 1e9 + 7;
const long long N = 300050;
long long n, a[N], dis[N], val[N], nodeans[N], ans = 0;
vector<vector<long long> > adj(N);
vector<long long> vis(N);
void dfs(long long v) {
  vis[v] = 1;
  val[v] += a[v];
  for (long long u : adj[v]) {
    if (!vis[u]) {
      dis[u] = dis[v] + 1;
      dfs(u);
      val[v] += val[u];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long test = 1;
  while (test--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1);
    for (long long i = 1; i <= n; i++) {
      ans += dis[i] * a[i];
      vis[i] = 0;
    }
    nodeans[1] = ans;
    long long tot = val[1];
    queue<long long> q;
    q.push(1);
    while (q.size()) {
      long long v = q.front();
      vis[v] = 1;
      q.pop();
      for (long long u : adj[v]) {
        if (!vis[u]) {
          nodeans[u] = nodeans[v] + (tot - 2 * val[u]);
          ans = max(ans, nodeans[u]);
          q.push(u);
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```