#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
vector<vector<long long> > v(1000005);
long long val[1000005], in[1000005], ans[1000005], total;
long long dfs(long long cur, long long par, long long level) {
  long long val1 = val[cur];
  ans[1] += level * val[cur];
  for (long long i = 0; i < v[cur].size(); i++) {
    if (v[cur][i] == par) continue;
    val1 += dfs(v[cur][i], cur, level + 1);
  }
  in[cur] = val1;
  return val1;
}
void ddfs(long long cur, long long par) {
  for (long long i = 0; i < v[cur].size(); i++) {
    if (v[cur][i] == par) continue;
    ans[v[cur][i]] = total - 2 * in[v[cur][i]] + ans[cur];
    ddfs(v[cur][i], cur);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) {
    cin >> val[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  total = dfs(1, 0, 0);
  ddfs(1, 0);
  long long finalAns = 0;
  for (long long i = 1; i <= n; i++) {
    finalAns = max(finalAns, ans[i]);
  }
  cout << finalAns << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize("O3")
using namespace std;
vector<vector<long long> > v(1000005);
long long val[1000005], in[1000005], ans[1000005], total;
long long dfs(long long cur, long long par, long long level) {
  long long val1 = val[cur];
  ans[1] += level * val[cur];
  for (long long i = 0; i < v[cur].size(); i++) {
    if (v[cur][i] == par) continue;
    val1 += dfs(v[cur][i], cur, level + 1);
  }
  in[cur] = val1;
  return val1;
}
void ddfs(long long cur, long long par) {
  for (long long i = 0; i < v[cur].size(); i++) {
    if (v[cur][i] == par) continue;
    ans[v[cur][i]] = total - 2 * in[v[cur][i]] + ans[cur];
    ddfs(v[cur][i], cur);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) {
    cin >> val[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  total = dfs(1, 0, 0);
  ddfs(1, 0);
  long long finalAns = 0;
  for (long long i = 1; i <= n; i++) {
    finalAns = max(finalAns, ans[i]);
  }
  cout << finalAns << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) solve();
  return 0;
}

```