#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
const long long N = 2e5 + 10;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto& x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
long long n, k, a[N], b, u, v;
long long sz[N];
long long in[N];
long long asz[N], sum = 0;
vector<long long> adj[N];
void dfs(long long u, long long p) {
  for (auto v : adj[u]) {
    if (v != p) {
      dfs(v, u);
      asz[u] += asz[v];
      in[u] += in[v] + asz[v];
    }
  }
}
void dfs1(long long u, long long p, long long n) {
  for (auto v : adj[u]) {
    if (v != p) {
      in[v] = in[u] - asz[v] + (sum - asz[v]);
      dfs1(v, u, n);
    }
  }
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    sz[i] = 1;
    sum += a[i];
    asz[i] = a[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  dfs1(1, 0, n);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans = max(ans, in[i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  long long t1 = 1;
  while (true) {
    solve();
    t1++;
    if (t1 > T) break;
  }
}

### Prompt

Please provide a Cpp coded solution 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 long long mod = 1e9 + 9;
const long long N = 2e5 + 10;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto& x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
long long n, k, a[N], b, u, v;
long long sz[N];
long long in[N];
long long asz[N], sum = 0;
vector<long long> adj[N];
void dfs(long long u, long long p) {
  for (auto v : adj[u]) {
    if (v != p) {
      dfs(v, u);
      asz[u] += asz[v];
      in[u] += in[v] + asz[v];
    }
  }
}
void dfs1(long long u, long long p, long long n) {
  for (auto v : adj[u]) {
    if (v != p) {
      in[v] = in[u] - asz[v] + (sum - asz[v]);
      dfs1(v, u, n);
    }
  }
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    sz[i] = 1;
    sum += a[i];
    asz[i] = a[i];
  }
  for (long long i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  dfs1(1, 0, n);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans = max(ans, in[i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  long long t1 = 1;
  while (true) {
    solve();
    t1++;
    if (t1 > T) break;
  }
}

```