#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
vector<long long int> a(N, -1);
vector<int> adj[N];
vector<long long int> h(N, -1);
vector<bool> visited(N, false);
vector<long long int> ans(N, 0);
vector<long long int> sum(N, 0);
long long int ful = 0;
long long int dfs(int v, int d = 0) {
  visited[v] = true;
  sum[v] += a[v];
  h[v] = d;
  for (int i = 0; i < adj[v].size(); i++) {
    if (visited[adj[v][i]] == false) sum[v] += dfs(adj[v][i], d + 1);
  }
  return sum[v];
}
void dfs2(int v, int par = -1) {
  visited[v] = true;
  if (par != -1) {
    ans[v] = ans[par] + ful - 2 * sum[v];
  }
  for (int i = 0; i < adj[v].size(); i++) {
    if (visited[adj[v][i]] == false) dfs2(adj[v][i], v);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  for (int i = 1; i <= n; i++) {
    ful += a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1);
  long long int xd = 0;
  for (int i = 2; i <= n; i++) {
    xd += a[i] * h[i];
  }
  ans[1] = xd;
  for (int i = 1; i <= n; i++) visited[i] = false;
  dfs2(1);
  long long int wyn = 0;
  for (int i = 1; i <= n; i++) wyn = max(wyn, ans[i]);
  cout << wyn;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int N = 2e5 + 5;
vector<long long int> a(N, -1);
vector<int> adj[N];
vector<long long int> h(N, -1);
vector<bool> visited(N, false);
vector<long long int> ans(N, 0);
vector<long long int> sum(N, 0);
long long int ful = 0;
long long int dfs(int v, int d = 0) {
  visited[v] = true;
  sum[v] += a[v];
  h[v] = d;
  for (int i = 0; i < adj[v].size(); i++) {
    if (visited[adj[v][i]] == false) sum[v] += dfs(adj[v][i], d + 1);
  }
  return sum[v];
}
void dfs2(int v, int par = -1) {
  visited[v] = true;
  if (par != -1) {
    ans[v] = ans[par] + ful - 2 * sum[v];
  }
  for (int i = 0; i < adj[v].size(); i++) {
    if (visited[adj[v][i]] == false) dfs2(adj[v][i], v);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  for (int i = 1; i <= n; i++) {
    ful += a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1);
  long long int xd = 0;
  for (int i = 2; i <= n; i++) {
    xd += a[i] * h[i];
  }
  ans[1] = xd;
  for (int i = 1; i <= n; i++) visited[i] = false;
  dfs2(1);
  long long int wyn = 0;
  for (int i = 1; i <= n; i++) wyn = max(wyn, ans[i]);
  cout << wyn;
  return 0;
}

```