#include <bits/stdc++.h>
const long long Inf = 1e18;
long long LINF = (long long)2e18;
using namespace std;
long long mod = 1e9 + 7;
long long mod2 = 1e9 + 123;
int n;
vector<long long> a, val;
vector<int> used;
vector<vector<int>> gr;
long long sum = 0;
long long cur = 0;
long long ans = 0;
long long dfs(int v, int d) {
  val[v] += a[v];
  cur += d * a[v];
  for (auto to : gr[v]) {
    if (val[to] == 0) {
      val[v] += dfs(to, d + 1);
    }
  }
  return val[v];
}
void dfs2(int v, long long tmp) {
  used[v] = 1;
  for (auto to : gr[v]) {
    if (!used[to]) {
      long long qq = tmp - val[to] + (sum - val[to]);
      ans = max(ans, qq);
      dfs2(to, qq);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a.resize(n);
  used.resize(n, 0);
  val.resize(n);
  gr.resize(n);
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int(i) = 0; (i) < n - 1; (i)++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  dfs(0, 0);
  ans = max(ans, cur);
  dfs2(0, ans);
  cout << ans;
  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>
const long long Inf = 1e18;
long long LINF = (long long)2e18;
using namespace std;
long long mod = 1e9 + 7;
long long mod2 = 1e9 + 123;
int n;
vector<long long> a, val;
vector<int> used;
vector<vector<int>> gr;
long long sum = 0;
long long cur = 0;
long long ans = 0;
long long dfs(int v, int d) {
  val[v] += a[v];
  cur += d * a[v];
  for (auto to : gr[v]) {
    if (val[to] == 0) {
      val[v] += dfs(to, d + 1);
    }
  }
  return val[v];
}
void dfs2(int v, long long tmp) {
  used[v] = 1;
  for (auto to : gr[v]) {
    if (!used[to]) {
      long long qq = tmp - val[to] + (sum - val[to]);
      ans = max(ans, qq);
      dfs2(to, qq);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a.resize(n);
  used.resize(n, 0);
  val.resize(n);
  gr.resize(n);
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int(i) = 0; (i) < n - 1; (i)++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  dfs(0, 0);
  ans = max(ans, cur);
  dfs2(0, ans);
  cout << ans;
  return 0;
}

```