#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ", ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << " => " << i.second << ')' << ' ';
  return os;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 2e5 + 5;
vector<vector<int>> v(N);
vector<int> val(N);
vector<long long> sum(N);
long long t, ans;
int n;
void dfs(int a, int par = -1, int dis = 0) {
  t += 1LL * dis * val[a];
  sum[a] = val[a];
  for (int i : v[a]) {
    if (i == par) continue;
    dfs(i, a, dis + 1);
    sum[a] += sum[i];
  }
}
void dfs2(int a, int par = -1) {
  ans = max(ans, t);
  for (int i : v[a]) {
    if (i == par) continue;
    t -= sum[i];
    sum[a] -= sum[i];
    t += sum[a];
    sum[i] += sum[a];
    dfs2(i, a);
    sum[i] -= sum[a];
    t -= sum[a];
    sum[a] += sum[i];
    t += sum[i];
  }
}
void solve() {
  dfs(1);
  dfs2(1);
  cout << ans << "\n";
}
int main() {
  fast();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  solve();
}

### 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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ", ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << " => " << i.second << ')' << ' ';
  return os;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 2e5 + 5;
vector<vector<int>> v(N);
vector<int> val(N);
vector<long long> sum(N);
long long t, ans;
int n;
void dfs(int a, int par = -1, int dis = 0) {
  t += 1LL * dis * val[a];
  sum[a] = val[a];
  for (int i : v[a]) {
    if (i == par) continue;
    dfs(i, a, dis + 1);
    sum[a] += sum[i];
  }
}
void dfs2(int a, int par = -1) {
  ans = max(ans, t);
  for (int i : v[a]) {
    if (i == par) continue;
    t -= sum[i];
    sum[a] -= sum[i];
    t += sum[a];
    sum[i] += sum[a];
    dfs2(i, a);
    sum[i] -= sum[a];
    t -= sum[a];
    sum[a] += sum[i];
    t += sum[i];
  }
}
void solve() {
  dfs(1);
  dfs2(1);
  cout << ans << "\n";
}
int main() {
  fast();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  solve();
}

```