#include <bits/stdc++.h>
using namespace std;
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
long long dfs1(vector<long long> adj[], long long dp[], long long ar[],
               long long u, long long par, long long height) {
  long long ans = ar[u] * height;
  dp[u] = ar[u];
  for (long long x : adj[u]) {
    if (x != par) {
      ans += dfs1(adj, dp, ar, x, u, height + 1);
      dp[u] += dp[x];
    }
  }
  return ans;
}
long long ans, ans1;
void dfs2(vector<long long> adj[], long long dp[], long long u, long long par) {
  ans = max(ans, ans1);
  for (long long x : adj[u]) {
    if (x != par) {
      ans1 -= dp[x];
      dp[u] -= dp[x];
      ans1 += dp[u];
      dp[x] += dp[u];
      dfs2(adj, dp, x, u);
      dp[x] -= dp[u];
      ans1 -= dp[u];
      dp[u] += dp[x];
      ans1 += dp[x];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long ar[n];
  for (int i = 0; i < n; i++) cin >> ar[i];
  vector<long long> adj[n];
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u - 1].emplace_back(v - 1);
    adj[v - 1].emplace_back(u - 1);
  }
  long long dp[n];
  ans1 = dfs1(adj, dp, ar, 0, -1, 0);
  dfs2(adj, dp, 0, -1);
  cout << ans << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
long long dfs1(vector<long long> adj[], long long dp[], long long ar[],
               long long u, long long par, long long height) {
  long long ans = ar[u] * height;
  dp[u] = ar[u];
  for (long long x : adj[u]) {
    if (x != par) {
      ans += dfs1(adj, dp, ar, x, u, height + 1);
      dp[u] += dp[x];
    }
  }
  return ans;
}
long long ans, ans1;
void dfs2(vector<long long> adj[], long long dp[], long long u, long long par) {
  ans = max(ans, ans1);
  for (long long x : adj[u]) {
    if (x != par) {
      ans1 -= dp[x];
      dp[u] -= dp[x];
      ans1 += dp[u];
      dp[x] += dp[u];
      dfs2(adj, dp, x, u);
      dp[x] -= dp[u];
      ans1 -= dp[u];
      dp[u] += dp[x];
      ans1 += dp[x];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long ar[n];
  for (int i = 0; i < n; i++) cin >> ar[i];
  vector<long long> adj[n];
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u - 1].emplace_back(v - 1);
    adj[v - 1].emplace_back(u - 1);
  }
  long long dp[n];
  ans1 = dfs1(adj, dp, ar, 0, -1, 0);
  dfs2(adj, dp, 0, -1);
  cout << ans << "\n";
  return 0;
}

```