#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 5;
const long long int inf64 = 1e18 + 5;
const int MAX = 2e5 + 5;
vector<int> adj[MAX];
long long int arr[MAX], dp[2][MAX], ans[MAX];
void dfs1(int x, int p) {
  dp[0][x] = dp[1][x] = arr[x];
  for (int v : adj[x]) {
    if (v != p) {
      dfs1(v, x);
      dp[0][x] += dp[0][v];
      dp[1][x] += dp[1][v];
    }
  }
}
void dfs2(int x, int p) {
  for (int v : adj[x]) {
    if (v != p) {
      dfs2(v, x);
      dp[1][x] += dp[1][v];
    }
  }
}
void dfs3(int x, int p, long long int res) {
  ans[x] = dp[1][x] + res + dp[0][1] - dp[0][x];
  for (int v : adj[x]) {
    if (v != p) dfs3(v, x, ans[x] - (dp[0][v] + dp[1][v]));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i, a, b;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> arr[i];
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs1(1, -1);
  dfs2(1, -1);
  for (i = 1; i <= n; i++) dp[1][i] -= dp[0][i];
  dfs3(1, -1, 0);
  cout << *max_element(ans + 1, ans + n + 1);
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 1e9 + 5;
const long long int inf64 = 1e18 + 5;
const int MAX = 2e5 + 5;
vector<int> adj[MAX];
long long int arr[MAX], dp[2][MAX], ans[MAX];
void dfs1(int x, int p) {
  dp[0][x] = dp[1][x] = arr[x];
  for (int v : adj[x]) {
    if (v != p) {
      dfs1(v, x);
      dp[0][x] += dp[0][v];
      dp[1][x] += dp[1][v];
    }
  }
}
void dfs2(int x, int p) {
  for (int v : adj[x]) {
    if (v != p) {
      dfs2(v, x);
      dp[1][x] += dp[1][v];
    }
  }
}
void dfs3(int x, int p, long long int res) {
  ans[x] = dp[1][x] + res + dp[0][1] - dp[0][x];
  for (int v : adj[x]) {
    if (v != p) dfs3(v, x, ans[x] - (dp[0][v] + dp[1][v]));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i, a, b;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> arr[i];
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs1(1, -1);
  dfs2(1, -1);
  for (i = 1; i <= n; i++) dp[1][i] -= dp[0][i];
  dfs3(1, -1, 0);
  cout << *max_element(ans + 1, ans + n + 1);
}

```