#include <bits/stdc++.h>
using namespace std;
long long res, sum;
void dfs(bool b[], vector<long long> adj[], long long levels[], long long u,
         long long sums[], long long a[]) {
  b[u] = true;
  sums[u] = a[u];
  vector<long long>::iterator it;
  for (it = adj[u].begin(); it != adj[u].end(); it++) {
    if (!b[(*it)]) {
      levels[(*it)] = levels[u] + 1;
      dfs(b, adj, levels, (*it), sums, a);
      sums[u] += sums[(*it)];
    }
  }
}
void dfs2(vector<long long> adj[], bool b[], long long sc, long long u,
          long long sums[]) {
  res = max(res, sc);
  b[u] = true;
  vector<long long>::iterator it;
  for (it = adj[u].begin(); it != adj[u].end(); it++) {
    if (!b[(*it)]) {
      long long temp = sc - 2 * sums[(*it)] + sum;
      dfs2(adj, b, temp, (*it), sums);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  long long *a = new long long[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<long long> adj[n + 1];
  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);
  }
  long long levels[n + 1];
  bool b[n + 1];
  for (long long i = 1; i <= n; i++) {
    b[i] = false;
  }
  levels[1] = 0;
  long long sums[n + 1];
  dfs(b, adj, levels, 1, sums, a);
  for (long long i = 1; i <= n; i++) {
    res += levels[i] * a[i];
    sum += a[i];
  }
  for (long long i = 1; i <= n; i++) {
    b[i] = false;
  }
  long long sc = res;
  dfs2(adj, b, sc, 1, sums);
  cout << res << "\n";
}

### Prompt

Your challenge is to write 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 res, sum;
void dfs(bool b[], vector<long long> adj[], long long levels[], long long u,
         long long sums[], long long a[]) {
  b[u] = true;
  sums[u] = a[u];
  vector<long long>::iterator it;
  for (it = adj[u].begin(); it != adj[u].end(); it++) {
    if (!b[(*it)]) {
      levels[(*it)] = levels[u] + 1;
      dfs(b, adj, levels, (*it), sums, a);
      sums[u] += sums[(*it)];
    }
  }
}
void dfs2(vector<long long> adj[], bool b[], long long sc, long long u,
          long long sums[]) {
  res = max(res, sc);
  b[u] = true;
  vector<long long>::iterator it;
  for (it = adj[u].begin(); it != adj[u].end(); it++) {
    if (!b[(*it)]) {
      long long temp = sc - 2 * sums[(*it)] + sum;
      dfs2(adj, b, temp, (*it), sums);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  long long *a = new long long[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<long long> adj[n + 1];
  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);
  }
  long long levels[n + 1];
  bool b[n + 1];
  for (long long i = 1; i <= n; i++) {
    b[i] = false;
  }
  levels[1] = 0;
  long long sums[n + 1];
  dfs(b, adj, levels, 1, sums, a);
  for (long long i = 1; i <= n; i++) {
    res += levels[i] * a[i];
    sum += a[i];
  }
  for (long long i = 1; i <= n; i++) {
    b[i] = false;
  }
  long long sc = res;
  dfs2(adj, b, sc, 1, sums);
  cout << res << "\n";
}

```