#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.14159265359;
const long long MOD = (long long)998244353ll;
const long long MAXN = (long long)2e5 + 10;
const long long INF = (long long)2242545357980376863;
const long double EPS = (long double)1e-8;
long long su = 0;
long long a[MAXN], dp1[MAXN], dp2[MAXN], dp3[MAXN];
vector<long long> G[MAXN];
void DFS(long long u, long long p) {
  dp1[u] = 0;
  dp2[u] = 0;
  for (auto adj : G[u]) {
    if (adj == p) continue;
    DFS(adj, u);
    dp1[u] += dp1[adj];
    dp2[u] += dp2[adj];
  }
  dp2[u] += dp1[u];
  dp1[u] += a[u];
}
void DFS2(long long u, long long p, long long v) {
  dp3[u] = v;
  long long tot = 0;
  for (auto adj : G[u]) {
    if (adj == p) continue;
    DFS2(adj, u, (v + dp2[u] - (dp2[adj] + dp1[adj])) + (su - dp1[adj]));
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) su += a[i];
  long long u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  DFS(1, -1);
  DFS2(1, -1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = max(ans, dp2[i] + dp3[i]);
  cout << ans;
  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;
const long double PI = 3.14159265359;
const long long MOD = (long long)998244353ll;
const long long MAXN = (long long)2e5 + 10;
const long long INF = (long long)2242545357980376863;
const long double EPS = (long double)1e-8;
long long su = 0;
long long a[MAXN], dp1[MAXN], dp2[MAXN], dp3[MAXN];
vector<long long> G[MAXN];
void DFS(long long u, long long p) {
  dp1[u] = 0;
  dp2[u] = 0;
  for (auto adj : G[u]) {
    if (adj == p) continue;
    DFS(adj, u);
    dp1[u] += dp1[adj];
    dp2[u] += dp2[adj];
  }
  dp2[u] += dp1[u];
  dp1[u] += a[u];
}
void DFS2(long long u, long long p, long long v) {
  dp3[u] = v;
  long long tot = 0;
  for (auto adj : G[u]) {
    if (adj == p) continue;
    DFS2(adj, u, (v + dp2[u] - (dp2[adj] + dp1[adj])) + (su - dp1[adj]));
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) su += a[i];
  long long u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  DFS(1, -1);
  DFS2(1, -1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = max(ans, dp2[i] + dp3[i]);
  cout << ans;
  return 0;
}

```