#include <bits/stdc++.h>
using namespace std;
typedef long double ld;
const long double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long maxn = 2e5 + 5;
long long a[maxn], dp[maxn], subtree[maxn], n;
vector<long long> g[maxn];
void dfs(long long x, long long p = -1) {
  for (auto i : g[x]) {
    if (i != p) {
      dfs(i, x);
      subtree[x] += subtree[i];
      dp[x] += dp[i];
    }
  }
  dp[x] += subtree[x];
  subtree[x] += a[x];
}
long long ans = 0, sum = 0;
void dfs2(long long x, long long p = -1) {
  if (p != -1) dp[x] = dp[p] + sum - 2 * subtree[x];
  ans = max(ans, dp[x]);
  for (auto i : g[x]) {
    if (i != p) dfs2(i, x);
  }
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  sum = subtree[0];
  dfs2(0);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
typedef long double ld;
const long double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long maxn = 2e5 + 5;
long long a[maxn], dp[maxn], subtree[maxn], n;
vector<long long> g[maxn];
void dfs(long long x, long long p = -1) {
  for (auto i : g[x]) {
    if (i != p) {
      dfs(i, x);
      subtree[x] += subtree[i];
      dp[x] += dp[i];
    }
  }
  dp[x] += subtree[x];
  subtree[x] += a[x];
}
long long ans = 0, sum = 0;
void dfs2(long long x, long long p = -1) {
  if (p != -1) dp[x] = dp[p] + sum - 2 * subtree[x];
  ans = max(ans, dp[x]);
  for (auto i : g[x]) {
    if (i != p) dfs2(i, x);
  }
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  sum = subtree[0];
  dfs2(0);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```