#include <bits/stdc++.h>
using LL = long long;
using LD = long double;
using namespace std;
using Graph = vector<vector<int>>;
Graph tree;
vector<LL> cost, DP, delta;
vector<bool> visited;
void dfs1(int u) {
  visited[u] = true;
  delta[u] = cost[u];
  for (int v : tree[u]) {
    if (!visited[v]) {
      dfs1(v);
      DP[u] += DP[v] + delta[v];
      delta[u] += delta[v];
    }
  }
}
LL max_cost = 0;
void dfs2(int u) {
  visited[u] = true;
  max_cost = max(max_cost, DP[u]);
  for (int v : tree[u]) {
    if (!visited[v]) {
      DP[v] = DP[u] + delta[0] - 2 * delta[v];
      dfs2(v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  cost.resize(n);
  for (auto& c : cost) cin >> c;
  tree.resize(n);
  for (auto _ = decltype(n - 1){0}; _ < (n - 1); ++_) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  DP.resize(n), delta.resize(n), visited.resize(n);
  dfs1(0);
  fill(std::begin(visited), std::end(visited), false);
  dfs2(0);
  cout << max_cost << endl;
}

### 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 LL = long long;
using LD = long double;
using namespace std;
using Graph = vector<vector<int>>;
Graph tree;
vector<LL> cost, DP, delta;
vector<bool> visited;
void dfs1(int u) {
  visited[u] = true;
  delta[u] = cost[u];
  for (int v : tree[u]) {
    if (!visited[v]) {
      dfs1(v);
      DP[u] += DP[v] + delta[v];
      delta[u] += delta[v];
    }
  }
}
LL max_cost = 0;
void dfs2(int u) {
  visited[u] = true;
  max_cost = max(max_cost, DP[u]);
  for (int v : tree[u]) {
    if (!visited[v]) {
      DP[v] = DP[u] + delta[0] - 2 * delta[v];
      dfs2(v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  cost.resize(n);
  for (auto& c : cost) cin >> c;
  tree.resize(n);
  for (auto _ = decltype(n - 1){0}; _ < (n - 1); ++_) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  DP.resize(n), delta.resize(n), visited.resize(n);
  dfs1(0);
  fill(std::begin(visited), std::end(visited), false);
  dfs2(0);
  cout << max_cost << endl;
}

```