#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 20;
long long a[N], now, ans, s[N];
vector<long long> gr[N];
void root(int v = 1, int h = 0, int par = -1) {
  now += h * a[v];
  s[v] = a[v];
  for (int u : gr[v])
    if (u != par) root(u, h + 1, v), s[v] += s[u];
}
void upd(int v = 1, int par = -1) {
  ans = max(ans, now);
  for (int u : gr[v]) {
    long long tmp = now;
    if (u != par) {
      now += s[1] - 2 * s[u], upd(u, v);
    }
    now = tmp;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    gr[--x].push_back(--y);
    gr[y].push_back(x);
  }
  root();
  upd();
  cout << ans;
}

### 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 = 2e6 + 20;
long long a[N], now, ans, s[N];
vector<long long> gr[N];
void root(int v = 1, int h = 0, int par = -1) {
  now += h * a[v];
  s[v] = a[v];
  for (int u : gr[v])
    if (u != par) root(u, h + 1, v), s[v] += s[u];
}
void upd(int v = 1, int par = -1) {
  ans = max(ans, now);
  for (int u : gr[v]) {
    long long tmp = now;
    if (u != par) {
      now += s[1] - 2 * s[u], upd(u, v);
    }
    now = tmp;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    gr[--x].push_back(--y);
    gr[y].push_back(x);
  }
  root();
  upd();
  cout << ans;
}

```