#include <bits/stdc++.h>
using namespace std;
const int MAX = 200005;
const int MOD = 998244353;
vector<int> g[MAX];
bool vis[MAX];
long long val[MAX], sum[MAX], in[MAX], out[MAX];
long long ans = 0;
void dfs_in(int n) {
  vis[n] = true;
  int child = 0;
  for (int node : g[n]) {
    if (!vis[node]) {
      child++;
      dfs_in(node);
      sum[n] += sum[node];
      in[n] += sum[node] + in[node];
    }
  }
  if (!child) in[n] = 0;
}
void dfs_out(int n, int par) {
  vis[n] = true;
  if (n == 1)
    out[n] = 0;
  else {
    long long a = in[par] - sum[n] - in[n];
    long long b = sum[1] - sum[n];
    out[n] = a + b;
  }
  in[n] += out[n];
  if (ans < in[n]) ans = in[n];
  for (int node : g[n])
    if (!vis[node]) dfs_out(node, n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> val[i];
    sum[i] = val[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs_in(1);
  ans = in[1];
  memset(vis, false, sizeof vis);
  dfs_out(1, 0);
  cout << ans << "\n";
}

### 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 MAX = 200005;
const int MOD = 998244353;
vector<int> g[MAX];
bool vis[MAX];
long long val[MAX], sum[MAX], in[MAX], out[MAX];
long long ans = 0;
void dfs_in(int n) {
  vis[n] = true;
  int child = 0;
  for (int node : g[n]) {
    if (!vis[node]) {
      child++;
      dfs_in(node);
      sum[n] += sum[node];
      in[n] += sum[node] + in[node];
    }
  }
  if (!child) in[n] = 0;
}
void dfs_out(int n, int par) {
  vis[n] = true;
  if (n == 1)
    out[n] = 0;
  else {
    long long a = in[par] - sum[n] - in[n];
    long long b = sum[1] - sum[n];
    out[n] = a + b;
  }
  in[n] += out[n];
  if (ans < in[n]) ans = in[n];
  for (int node : g[n])
    if (!vis[node]) dfs_out(node, n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> val[i];
    sum[i] = val[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs_in(1);
  ans = in[1];
  memset(vis, false, sizeof vis);
  dfs_out(1, 0);
  cout << ans << "\n";
}

```