#include <bits/stdc++.h>
using namespace std;
namespace solve {
long long int a[200007], sum[200007], dpDown[200007], dpUp[200007];
vector<int> graph[200007];
void rec(int u, int p) {
  sum[u] = a[u];
  for (int v : graph[u]) {
    if (v == p) continue;
    rec(v, u);
    dpDown[u] += dpDown[v] + sum[v];
    sum[u] += sum[v];
  }
}
void dfs(int u, int p) {
  for (int v : graph[u]) {
    if (v == p) continue;
    dpUp[v] = dpDown[u] - dpDown[v] - sum[v] + (sum[1] - sum[v]) + dpUp[u];
    dfs(v, u);
  }
}
void takeInput() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  rec(1, -1);
  dfs(1, -1);
  long long int res = 0;
  for (int i = 1; i <= n; i++) {
    res = max(res, dpDown[i] + dpUp[i]);
  }
  cout << res << endl;
}
}  // namespace solve
int main() {
  solve::takeInput();
  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;
namespace solve {
long long int a[200007], sum[200007], dpDown[200007], dpUp[200007];
vector<int> graph[200007];
void rec(int u, int p) {
  sum[u] = a[u];
  for (int v : graph[u]) {
    if (v == p) continue;
    rec(v, u);
    dpDown[u] += dpDown[v] + sum[v];
    sum[u] += sum[v];
  }
}
void dfs(int u, int p) {
  for (int v : graph[u]) {
    if (v == p) continue;
    dpUp[v] = dpDown[u] - dpDown[v] - sum[v] + (sum[1] - sum[v]) + dpUp[u];
    dfs(v, u);
  }
}
void takeInput() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  rec(1, -1);
  dfs(1, -1);
  long long int res = 0;
  for (int i = 1; i <= n; i++) {
    res = max(res, dpDown[i] + dpUp[i]);
  }
  cout << res << endl;
}
}  // namespace solve
int main() {
  solve::takeInput();
  return 0;
}

```