#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 100003;
const double EPSILON = 0.0000000001;
const int NMAX = 2e5 + 5;
vector<int> graph[NMAX];
int n;
int cost[NMAX];
long long res, ans;
long long sum[NMAX];
void dfs(int node, int father, int depth) {
  res += 1LL * depth * cost[node];
  sum[node] = cost[node];
  for (int next : graph[node])
    if (next != father) {
      dfs(next, node, depth + 1);
      sum[node] += sum[next];
    }
}
void extend(int node, int father) {
  ans = max(ans, res);
  for (int next : graph[node])
    if (next != father) {
      res -= sum[next];
      sum[node] -= sum[next];
      res += sum[node];
      sum[next] += sum[node];
      extend(next, node);
      sum[next] -= sum[node];
      res -= sum[node];
      sum[node] += sum[next];
      res += sum[next];
    }
}
void read() {
  int x, y;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> cost[i];
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
int main() {
  read();
  dfs(1, -1, 0);
  extend(1, -1);
  cout << ans;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 int INF = 0x3f3f3f3f;
const int MOD = 100003;
const double EPSILON = 0.0000000001;
const int NMAX = 2e5 + 5;
vector<int> graph[NMAX];
int n;
int cost[NMAX];
long long res, ans;
long long sum[NMAX];
void dfs(int node, int father, int depth) {
  res += 1LL * depth * cost[node];
  sum[node] = cost[node];
  for (int next : graph[node])
    if (next != father) {
      dfs(next, node, depth + 1);
      sum[node] += sum[next];
    }
}
void extend(int node, int father) {
  ans = max(ans, res);
  for (int next : graph[node])
    if (next != father) {
      res -= sum[next];
      sum[node] -= sum[next];
      res += sum[node];
      sum[next] += sum[node];
      extend(next, node);
      sum[next] -= sum[node];
      res -= sum[node];
      sum[node] += sum[next];
      res += sum[next];
    }
}
void read() {
  int x, y;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> cost[i];
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
int main() {
  read();
  dfs(1, -1, 0);
  extend(1, -1);
  cout << ans;
  return 0;
}

```