#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a((size_t)n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long>> adj((size_t)n);
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long sum = 0;
  vector<long long> sub((size_t)n, 0);
  function<void(long long, long long, long long)> visit_1 =
      [&](long long u, long long v, long long d) {
        sum += a[v] * d;
        sub[v] += a[v];
        for (long long w : adj[v]) {
          if (w != u) {
            visit_1(v, w, d + 1);
            sub[v] += sub[w];
          }
        }
      };
  visit_1(-1, 0, 0);
  long long all = accumulate(a.begin(), a.end(), (long long)0);
  long long ans = 0;
  function<void(long long, long long)> visit_2 = [&](long long u, long long v) {
    ans = max(ans, sum);
    for (long long w : adj[v]) {
      if (w != u) {
        sum += all - sub[w];
        sum -= sub[w];
        visit_2(v, w);
        sum += sub[w];
        sum -= all - sub[w];
      }
    }
  };
  visit_2(-1, 0);
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a((size_t)n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long>> adj((size_t)n);
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long sum = 0;
  vector<long long> sub((size_t)n, 0);
  function<void(long long, long long, long long)> visit_1 =
      [&](long long u, long long v, long long d) {
        sum += a[v] * d;
        sub[v] += a[v];
        for (long long w : adj[v]) {
          if (w != u) {
            visit_1(v, w, d + 1);
            sub[v] += sub[w];
          }
        }
      };
  visit_1(-1, 0, 0);
  long long all = accumulate(a.begin(), a.end(), (long long)0);
  long long ans = 0;
  function<void(long long, long long)> visit_2 = [&](long long u, long long v) {
    ans = max(ans, sum);
    for (long long w : adj[v]) {
      if (w != u) {
        sum += all - sub[w];
        sum -= sub[w];
        visit_2(v, w);
        sum += sub[w];
        sum -= all - sub[w];
      }
    }
  };
  visit_2(-1, 0);
  cout << ans << endl;
  return 0;
}

```