#include <bits/stdc++.h>
using namespace std;
void setIO(const string &name) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin.exceptions(istream::failbit);
}
const int inf = 0x3f3f3f3f, mod = 1e9 + 7, maxn = 2e5 + 5;
pair<long long, long long> down[maxn], up[maxn];
int a[maxn];
vector<int> adj[maxn];
void dfs1(int c, int p) {
  for (auto &i : adj[c]) {
    if (i != p) {
      dfs1(i, c);
      down[c].first += down[i].first + down[i].second;
      down[c].second += down[i].second;
    }
  }
  down[c].second += a[c];
}
void dfs2(int c, int p) {
  if (p != -1) {
    up[c].first += up[p].first + down[p].first - down[c].first - down[c].second;
    up[c].first += up[p].second + down[p].second - down[c].second;
    up[c].second += up[p].second + down[p].second - down[c].second;
  }
  for (auto &i : adj[c]) {
    if (i != p) {
      dfs2(i, c);
    }
  }
}
int main() {
  setIO("1");
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    int b, c;
    cin >> b >> c;
    --b, --c;
    adj[b].push_back(c);
    adj[c].push_back(b);
  }
  dfs1(0, -1);
  dfs2(0, -1);
  long long sol = 0;
  for (int i = 0; i < n; ++i) {
    sol = max(sol, down[i].first + up[i].first);
  }
  cout << sol << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
void setIO(const string &name) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin.exceptions(istream::failbit);
}
const int inf = 0x3f3f3f3f, mod = 1e9 + 7, maxn = 2e5 + 5;
pair<long long, long long> down[maxn], up[maxn];
int a[maxn];
vector<int> adj[maxn];
void dfs1(int c, int p) {
  for (auto &i : adj[c]) {
    if (i != p) {
      dfs1(i, c);
      down[c].first += down[i].first + down[i].second;
      down[c].second += down[i].second;
    }
  }
  down[c].second += a[c];
}
void dfs2(int c, int p) {
  if (p != -1) {
    up[c].first += up[p].first + down[p].first - down[c].first - down[c].second;
    up[c].first += up[p].second + down[p].second - down[c].second;
    up[c].second += up[p].second + down[p].second - down[c].second;
  }
  for (auto &i : adj[c]) {
    if (i != p) {
      dfs2(i, c);
    }
  }
}
int main() {
  setIO("1");
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    int b, c;
    cin >> b >> c;
    --b, --c;
    adj[b].push_back(c);
    adj[c].push_back(b);
  }
  dfs1(0, -1);
  dfs2(0, -1);
  long long sol = 0;
  for (int i = 0; i < n; ++i) {
    sol = max(sol, down[i].first + up[i].first);
  }
  cout << sol << "\n";
  return 0;
}

```