#include <bits/stdc++.h>
using namespace std;
void dfs(long long v, vector<vector<long long> >& g, vector<long long>& d,
         vector<long long>& mark, vector<long long>& a,
         vector<long long>& siz) {
  mark[v] = 1;
  siz[v] += a[v];
  for (auto u : g[v]) {
    if (!mark[u]) {
      d[u] = d[v] + 1;
      dfs(u, g, d, mark, a, siz);
      siz[v] += siz[u];
    }
  }
}
void bfs(long long s, vector<vector<long long> >& g, vector<long long>& siz,
         vector<long long>& aa, vector<long long>& a, vector<long long>& mark) {
  long long n = g.size();
  mark.assign(n, 0);
  long long sum = 0;
  for (auto u : a) sum += u;
  mark[s] = 1;
  queue<long long> q;
  q.push(s);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    for (auto u : g[v]) {
      if (!mark[u]) {
        mark[u] = 1;
        aa[u] = aa[v] + sum - 2 * siz[u];
        q.push(u);
      }
    }
  }
}
int main() {
  long long n;
  cin >> n;
  vector<vector<long long> > g(n);
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long v, u;
  for (long long i = 0; i < n - 1; i++) {
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  vector<long long> d(n), mark(n), siz(n, 0);
  d.assign(n, 0);
  mark.assign(n, 0);
  siz.assign(n, 0);
  dfs(0, g, d, mark, a, siz);
  vector<long long> aa(n);
  aa.assign(n, 0);
  for (long long i = 0; i < n; i++) aa[0] += (d[i] * a[i]);
  bfs(0, g, siz, aa, a, mark);
  long long max = 0;
  for (long long i = 0; i < n; i++)
    if (aa[i] > max) max = aa[i];
  cout << max << "\n";
  return 0;
}

### Prompt

Generate a cpp solution 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 dfs(long long v, vector<vector<long long> >& g, vector<long long>& d,
         vector<long long>& mark, vector<long long>& a,
         vector<long long>& siz) {
  mark[v] = 1;
  siz[v] += a[v];
  for (auto u : g[v]) {
    if (!mark[u]) {
      d[u] = d[v] + 1;
      dfs(u, g, d, mark, a, siz);
      siz[v] += siz[u];
    }
  }
}
void bfs(long long s, vector<vector<long long> >& g, vector<long long>& siz,
         vector<long long>& aa, vector<long long>& a, vector<long long>& mark) {
  long long n = g.size();
  mark.assign(n, 0);
  long long sum = 0;
  for (auto u : a) sum += u;
  mark[s] = 1;
  queue<long long> q;
  q.push(s);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    for (auto u : g[v]) {
      if (!mark[u]) {
        mark[u] = 1;
        aa[u] = aa[v] + sum - 2 * siz[u];
        q.push(u);
      }
    }
  }
}
int main() {
  long long n;
  cin >> n;
  vector<vector<long long> > g(n);
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long v, u;
  for (long long i = 0; i < n - 1; i++) {
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  vector<long long> d(n), mark(n), siz(n, 0);
  d.assign(n, 0);
  mark.assign(n, 0);
  siz.assign(n, 0);
  dfs(0, g, d, mark, a, siz);
  vector<long long> aa(n);
  aa.assign(n, 0);
  for (long long i = 0; i < n; i++) aa[0] += (d[i] * a[i]);
  bfs(0, g, siz, aa, a, mark);
  long long max = 0;
  for (long long i = 0; i < n; i++)
    if (aa[i] > max) max = aa[i];
  cout << max << "\n";
  return 0;
}

```