#include <bits/stdc++.h>
using namespace std;
long long n;
vector<vector<long long> > g;
vector<long long> a, dp_down, sum, dp;
void dfs1(long long v, long long p = -1) {
  sum[v] = a[v];
  for (long long to : g[v]) {
    if (to == p) continue;
    dfs1(to, v);
    sum[v] += sum[to];
    dp_down[v] += dp_down[to] + sum[to];
  }
}
void dfs2(long long v, long long upper_dp, long long cur_sum,
          long long p = -1) {
  dp[v] = upper_dp + dp_down[v];
  for (long long to : g[v]) {
    if (to == p) continue;
    long long dp_sons = dp_down[v] - dp_down[to] - sum[to];
    long long sum_sons = sum[v] - sum[to];
    dfs2(to, upper_dp + dp_sons + cur_sum + sum_sons, cur_sum + sum_sons, v);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  a.resize(n);
  dp_down.assign(n, 0);
  sum.assign(n, 0);
  dp.assign(n, 0);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(0);
  dfs2(0, 0, 0);
  cout << *max_element((dp).begin(), (dp).end());
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long n;
vector<vector<long long> > g;
vector<long long> a, dp_down, sum, dp;
void dfs1(long long v, long long p = -1) {
  sum[v] = a[v];
  for (long long to : g[v]) {
    if (to == p) continue;
    dfs1(to, v);
    sum[v] += sum[to];
    dp_down[v] += dp_down[to] + sum[to];
  }
}
void dfs2(long long v, long long upper_dp, long long cur_sum,
          long long p = -1) {
  dp[v] = upper_dp + dp_down[v];
  for (long long to : g[v]) {
    if (to == p) continue;
    long long dp_sons = dp_down[v] - dp_down[to] - sum[to];
    long long sum_sons = sum[v] - sum[to];
    dfs2(to, upper_dp + dp_sons + cur_sum + sum_sons, cur_sum + sum_sons, v);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  a.resize(n);
  dp_down.assign(n, 0);
  sum.assign(n, 0);
  dp.assign(n, 0);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(0);
  dfs2(0, 0, 0);
  cout << *max_element((dp).begin(), (dp).end());
  return 0;
}

```