#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > adj;
vector<long long> total;
vector<long long> dp;
vector<long long> a;
vector<long long> ans;
void dfs(int x, int parent) {
  for (auto& s : adj[x]) {
    if (s == parent) continue;
    dfs(s, x);
    total[x] += a[s] + total[s];
    dp[x] += a[s] + total[s] + dp[s];
  }
}
long long something;
void dfs2(int x, int parent) {
  if (parent != -1) ans[x] = ans[parent] - 2 * (a[x] + total[x]) + something;
  for (auto& s : adj[x]) {
    if (s == parent) continue;
    dfs2(s, x);
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n;
  cin >> n;
  a.assign(n, 0);
  adj.assign(n, vector<long long>());
  total.assign(n, 0);
  dp.assign(n, 0);
  for (auto& s : a) cin >> s;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, -1);
  ans.assign(n, 0);
  ans[0] = dp[0];
  something = total[0] + a[0];
  dfs2(0, -1);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    s = max(s, ans[i]);
  }
  cout << s << "\n";
}

### Prompt

Your challenge is to write 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;
vector<vector<long long> > adj;
vector<long long> total;
vector<long long> dp;
vector<long long> a;
vector<long long> ans;
void dfs(int x, int parent) {
  for (auto& s : adj[x]) {
    if (s == parent) continue;
    dfs(s, x);
    total[x] += a[s] + total[s];
    dp[x] += a[s] + total[s] + dp[s];
  }
}
long long something;
void dfs2(int x, int parent) {
  if (parent != -1) ans[x] = ans[parent] - 2 * (a[x] + total[x]) + something;
  for (auto& s : adj[x]) {
    if (s == parent) continue;
    dfs2(s, x);
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n;
  cin >> n;
  a.assign(n, 0);
  adj.assign(n, vector<long long>());
  total.assign(n, 0);
  dp.assign(n, 0);
  for (auto& s : a) cin >> s;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, -1);
  ans.assign(n, 0);
  ans[0] = dp[0];
  something = total[0] + a[0];
  dfs2(0, -1);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    s = max(s, ans[i]);
  }
  cout << s << "\n";
}

```