#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long double PI = 3.14159265358979323846;
const long long MOD = 1e+9 + 7;
const long long INF = LLONG_MAX;
const int INFi = INT_MAX;
const long long N = 2e+5 + 8;
vector<long long> adj[N];
long long vis[N];
long long dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
long long test_cases = 1;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
;
long long n, a[N], u, v, res, ans;
long long dp[N], accum[N];
void dfs(long long v, long long d) {
  vis[v] = 1;
  accum[v] = a[v];
  res += d * a[v];
  for (auto to : adj[v]) {
    if (!vis[to]) {
      dfs(to, d + 1);
      accum[v] += accum[to];
    }
  }
  vis[v] = 0;
}
void dfs2(long long v) {
  ans = max(ans, dp[v]);
  vis[v] = 1;
  for (auto to : adj[v]) {
    if (!vis[to]) {
      accum[v] -= accum[to];
      dp[to] = dp[v] + accum[v] - accum[to];
      accum[to] += accum[v];
      dfs2(to);
      accum[to] -= accum[v];
      accum[v] += accum[to];
    }
  }
}
void MAIN(long long tc) {
  cin >> n;
  for (long long i = (1); i <= n; i++) cin >> 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);
  dp[1] = res;
  dfs2(1);
  cout << ans << " ";
  cout << "\n";
  ;
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (long long i = (1); i <= test_cases; i++) {
    MAIN(i);
  }
}

### 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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long double PI = 3.14159265358979323846;
const long long MOD = 1e+9 + 7;
const long long INF = LLONG_MAX;
const int INFi = INT_MAX;
const long long N = 2e+5 + 8;
vector<long long> adj[N];
long long vis[N];
long long dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
long long test_cases = 1;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
;
long long n, a[N], u, v, res, ans;
long long dp[N], accum[N];
void dfs(long long v, long long d) {
  vis[v] = 1;
  accum[v] = a[v];
  res += d * a[v];
  for (auto to : adj[v]) {
    if (!vis[to]) {
      dfs(to, d + 1);
      accum[v] += accum[to];
    }
  }
  vis[v] = 0;
}
void dfs2(long long v) {
  ans = max(ans, dp[v]);
  vis[v] = 1;
  for (auto to : adj[v]) {
    if (!vis[to]) {
      accum[v] -= accum[to];
      dp[to] = dp[v] + accum[v] - accum[to];
      accum[to] += accum[v];
      dfs2(to);
      accum[to] -= accum[v];
      accum[v] += accum[to];
    }
  }
}
void MAIN(long long tc) {
  cin >> n;
  for (long long i = (1); i <= n; i++) cin >> 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);
  dp[1] = res;
  dfs2(1);
  cout << ans << " ";
  cout << "\n";
  ;
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (long long i = (1); i <= test_cases; i++) {
    MAIN(i);
  }
}

```