#include <bits/stdc++.h>
using namespace std;
const long long N = 200005, OO = 1110111110111, mod = 1000000007,
                mod2 = 998244353;
int n, l, r;
int a[N], vis[N];
vector<int> adj[N];
long long sol, total, par, all;
long long sub[N];
long long calc(int x, int r) {
  vis[x] = 1;
  total += 1ll * r * a[x];
  long long ret = a[x];
  for (auto u : adj[x]) {
    if (vis[u]) continue;
    ret += calc(u, r + 1);
  }
  vis[x] = 0;
  return sub[x] = ret;
}
void root(int x, long long now) {
  vis[x] = 1;
  now -= sub[x];
  now += all - sub[x];
  sol = max(sol, now);
  for (auto u : adj[x]) {
    if (vis[u]) continue;
    root(u, now);
  }
  vis[x] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    all += a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &l, &r);
    adj[l].push_back(r);
    adj[r].push_back(l);
  }
  calc(1, 1);
  root(1, total);
  printf("%I64d", sol);
  return 0;
}

### 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 N = 200005, OO = 1110111110111, mod = 1000000007,
                mod2 = 998244353;
int n, l, r;
int a[N], vis[N];
vector<int> adj[N];
long long sol, total, par, all;
long long sub[N];
long long calc(int x, int r) {
  vis[x] = 1;
  total += 1ll * r * a[x];
  long long ret = a[x];
  for (auto u : adj[x]) {
    if (vis[u]) continue;
    ret += calc(u, r + 1);
  }
  vis[x] = 0;
  return sub[x] = ret;
}
void root(int x, long long now) {
  vis[x] = 1;
  now -= sub[x];
  now += all - sub[x];
  sol = max(sol, now);
  for (auto u : adj[x]) {
    if (vis[u]) continue;
    root(u, now);
  }
  vis[x] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    all += a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &l, &r);
    adj[l].push_back(r);
    adj[r].push_back(l);
  }
  calc(1, 1);
  root(1, total);
  printf("%I64d", sol);
  return 0;
}

```