#include <bits/stdc++.h>
using namespace std;
const int inf = 2e9 + 500;
const int N = 2e5 + 1;
const int bits = 20;
const long double pi = 3.14159265358979323846;
const long long mod = 998244353;
const long long p = 29;
int n;
vector<int> adj[N];
long long dp1[N], dp2[N], sub[N], arr[N];
void dfs1(int node, int par = 0) {
  sub[node] = arr[node];
  for (int nxt : adj[node]) {
    if (nxt != par) {
      dfs1(nxt, node);
      sub[node] += sub[nxt];
      dp1[node] += dp1[nxt] + sub[nxt];
    }
  }
}
void dfs2(int node, int par = 0) {
  if (par != 0) {
    dp2[node] = dp2[par] - sub[node] + sub[1] - sub[node];
  }
  for (int nxt : adj[node]) {
    if (nxt != par) {
      dfs2(nxt, node);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  std::cout << std::fixed << std::setprecision(20);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1);
  dp2[1] = dp1[1];
  dfs2(1);
  long long ans = dp2[1];
  for (int i = 2; i <= n; i++) {
    ans = max(ans, dp2[i]);
  }
  cout << ans << "\n";
}

### 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;
const int inf = 2e9 + 500;
const int N = 2e5 + 1;
const int bits = 20;
const long double pi = 3.14159265358979323846;
const long long mod = 998244353;
const long long p = 29;
int n;
vector<int> adj[N];
long long dp1[N], dp2[N], sub[N], arr[N];
void dfs1(int node, int par = 0) {
  sub[node] = arr[node];
  for (int nxt : adj[node]) {
    if (nxt != par) {
      dfs1(nxt, node);
      sub[node] += sub[nxt];
      dp1[node] += dp1[nxt] + sub[nxt];
    }
  }
}
void dfs2(int node, int par = 0) {
  if (par != 0) {
    dp2[node] = dp2[par] - sub[node] + sub[1] - sub[node];
  }
  for (int nxt : adj[node]) {
    if (nxt != par) {
      dfs2(nxt, node);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  std::cout << std::fixed << std::setprecision(20);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1);
  dp2[1] = dp1[1];
  dfs2(1);
  long long ans = dp2[1];
  for (int i = 2; i <= n; i++) {
    ans = max(ans, dp2[i]);
  }
  cout << ans << "\n";
}

```