#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 2e5 + 5;
long long res = 0, mx = 0;
long long sub[MAX_N];
int custo[MAX_N];
vector<vector<int> > g(MAX_N);
void dfs(int node, int p = -1, int d = 0) {
  res += 1LL * d * custo[node];
  sub[node] = custo[node];
  for (auto to : g[node]) {
    if (to != p) {
      dfs(to, node, d + 1);
      sub[node] += sub[to];
    }
  }
}
void root(int node, int p = -1) {
  mx = max(mx, res);
  for (auto to : g[node]) {
    if (to != p) {
      long long old_res = res;
      long long old_node = sub[node];
      long long old_to = sub[to];
      res -= sub[to];
      sub[node] -= sub[to];
      res += sub[node];
      sub[to] += sub[node];
      root(to, node);
      res = old_res;
      sub[node] = old_node;
      sub[to] = old_to;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> custo[i];
  for (int i = 0; i < n - 1; i++) {
    int st, et;
    cin >> st >> et;
    --st;
    --et;
    g[st].emplace_back(et);
    g[et].emplace_back(st);
  }
  dfs(0);
  root(0);
  cout << mx << '\n';
  return 0;
}

### Prompt

Please formulate 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;
const int MAX_N = 2e5 + 5;
long long res = 0, mx = 0;
long long sub[MAX_N];
int custo[MAX_N];
vector<vector<int> > g(MAX_N);
void dfs(int node, int p = -1, int d = 0) {
  res += 1LL * d * custo[node];
  sub[node] = custo[node];
  for (auto to : g[node]) {
    if (to != p) {
      dfs(to, node, d + 1);
      sub[node] += sub[to];
    }
  }
}
void root(int node, int p = -1) {
  mx = max(mx, res);
  for (auto to : g[node]) {
    if (to != p) {
      long long old_res = res;
      long long old_node = sub[node];
      long long old_to = sub[to];
      res -= sub[to];
      sub[node] -= sub[to];
      res += sub[node];
      sub[to] += sub[node];
      root(to, node);
      res = old_res;
      sub[node] = old_node;
      sub[to] = old_to;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> custo[i];
  for (int i = 0; i < n - 1; i++) {
    int st, et;
    cin >> st >> et;
    --st;
    --et;
    g[st].emplace_back(et);
    g[et].emplace_back(st);
  }
  dfs(0);
  root(0);
  cout << mx << '\n';
  return 0;
}

```