#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<long long int> dis(n);
  vector<long long int> ans(n);
  vector<bool> vis(n, 0);
  long long int acc = 0;
  for (long int i = 0; i < n; i++) {
    cin >> dis[i];
    acc += dis[i];
  }
  vector<vector<int>> edg(n);
  for (long int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edg[a].push_back(b);
    edg[b].push_back(a);
  }
  vector<vector<int>> p(3);
  p[0].push_back(0);
  p[1].push_back(0);
  p[2].push_back(0);
  vis[0] = 1;
  int i = 0;
  long long int x = 0;
  while (p[0].size() < n) {
    for (long int j = 0; j < edg[p[0][i]].size(); j++) {
      if (!vis[edg[p[0][i]][j]]) {
        p[0].push_back(edg[p[0][i]][j]);
        p[1].push_back(p[0][i]);
        p[2].push_back(p[2][i] + 1);
        x += ((long long int)(p[2][i] + 1)) * (dis[edg[p[0][i]][j]]);
        vis[edg[p[0][i]][j]] = 1;
      }
    }
    i++;
  }
  for (int j = n - 1; j > 0; j--) {
    dis[p[1][j]] += dis[p[0][j]];
  }
  ans[0] = x;
  for (long int i = 0; i < n; i++) {
    if (i == 0) continue;
    ans[p[0][i]] = acc + ans[p[1][i]] - 2 * dis[p[0][i]];
  }
  cout << *max_element(ans.begin(), ans.end());
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  int n;
  cin >> n;
  vector<long long int> dis(n);
  vector<long long int> ans(n);
  vector<bool> vis(n, 0);
  long long int acc = 0;
  for (long int i = 0; i < n; i++) {
    cin >> dis[i];
    acc += dis[i];
  }
  vector<vector<int>> edg(n);
  for (long int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edg[a].push_back(b);
    edg[b].push_back(a);
  }
  vector<vector<int>> p(3);
  p[0].push_back(0);
  p[1].push_back(0);
  p[2].push_back(0);
  vis[0] = 1;
  int i = 0;
  long long int x = 0;
  while (p[0].size() < n) {
    for (long int j = 0; j < edg[p[0][i]].size(); j++) {
      if (!vis[edg[p[0][i]][j]]) {
        p[0].push_back(edg[p[0][i]][j]);
        p[1].push_back(p[0][i]);
        p[2].push_back(p[2][i] + 1);
        x += ((long long int)(p[2][i] + 1)) * (dis[edg[p[0][i]][j]]);
        vis[edg[p[0][i]][j]] = 1;
      }
    }
    i++;
  }
  for (int j = n - 1; j > 0; j--) {
    dis[p[1][j]] += dis[p[0][j]];
  }
  ans[0] = x;
  for (long int i = 0; i < n; i++) {
    if (i == 0) continue;
    ans[p[0][i]] = acc + ans[p[1][i]] - 2 * dis[p[0][i]];
  }
  cout << *max_element(ans.begin(), ans.end());
  return 0;
}

```