#include <bits/stdc++.h>
using namespace std;
int n;
long long sol, total;
long long arr[200005], dp[200005], children[200005];
vector<int> graph[200005];
bitset<200005> check;
void dfs1(int now);
void dfs2(int now);
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    total += arr[i];
  }
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  dfs1(1);
  for (int i = 1; i <= n; ++i) {
    check[i] = false;
  }
  dfs2(1);
  for (int i = 1; i <= n; ++i) {
    sol = max(sol, dp[i]);
  }
  cout << sol;
  return 0;
}
void dfs1(int now) {
  check[now] = true;
  for (auto nxt : graph[now]) {
    if (check[nxt]) continue;
    dfs1(nxt);
    children[now] += (children[nxt] + arr[nxt]);
    dp[now] += (dp[nxt] + children[nxt] + arr[nxt]);
  }
}
void dfs2(int now) {
  check[now] = true;
  for (auto nxt : graph[now]) {
    if (check[nxt]) continue;
    dp[nxt] = dp[now] + total - 2 * (children[nxt] + arr[nxt]);
    dfs2(nxt);
  }
}

### Prompt

Your task is to create 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;
int n;
long long sol, total;
long long arr[200005], dp[200005], children[200005];
vector<int> graph[200005];
bitset<200005> check;
void dfs1(int now);
void dfs2(int now);
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    total += arr[i];
  }
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  dfs1(1);
  for (int i = 1; i <= n; ++i) {
    check[i] = false;
  }
  dfs2(1);
  for (int i = 1; i <= n; ++i) {
    sol = max(sol, dp[i]);
  }
  cout << sol;
  return 0;
}
void dfs1(int now) {
  check[now] = true;
  for (auto nxt : graph[now]) {
    if (check[nxt]) continue;
    dfs1(nxt);
    children[now] += (children[nxt] + arr[nxt]);
    dp[now] += (dp[nxt] + children[nxt] + arr[nxt]);
  }
}
void dfs2(int now) {
  check[now] = true;
  for (auto nxt : graph[now]) {
    if (check[nxt]) continue;
    dp[nxt] = dp[now] + total - 2 * (children[nxt] + arr[nxt]);
    dfs2(nxt);
  }
}

```