#include <bits/stdc++.h>
using namespace std;
const long long N = 2000 * 100 + 1000;
vector<long long> adj[N];
long long a[N], h[N], t[N];
bool mark[N];
long long f(long long p) {
  mark[p] = true;
  long long ans = h[p] * a[p];
  t[p] = a[p];
  for (auto q : adj[p])
    if (mark[q] == false) {
      h[q] = h[p] + 1;
      ans += f(q);
      t[p] += t[q];
    }
  return ans;
}
long long ans = 0, sum = 0;
void calc(long long val, long long p) {
  mark[p] = true;
  ans = max(ans, val);
  for (auto q : adj[p])
    if (mark[q] == false) {
      long long k = sum - 2 * t[q];
      calc(val + k, q);
    }
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  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 val = f(0);
  for (long long i = 0; i < N; i++) mark[i] = false;
  calc(val, 0);
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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;
const long long N = 2000 * 100 + 1000;
vector<long long> adj[N];
long long a[N], h[N], t[N];
bool mark[N];
long long f(long long p) {
  mark[p] = true;
  long long ans = h[p] * a[p];
  t[p] = a[p];
  for (auto q : adj[p])
    if (mark[q] == false) {
      h[q] = h[p] + 1;
      ans += f(q);
      t[p] += t[q];
    }
  return ans;
}
long long ans = 0, sum = 0;
void calc(long long val, long long p) {
  mark[p] = true;
  ans = max(ans, val);
  for (auto q : adj[p])
    if (mark[q] == false) {
      long long k = sum - 2 * t[q];
      calc(val + k, q);
    }
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  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 val = f(0);
  for (long long i = 0; i < N; i++) mark[i] = false;
  calc(val, 0);
  cout << ans << '\n';
  return 0;
}

```