#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 3;
const int maxm = 1e6 + 3;
const int mod = 1e9 + 7;
const int oo = 1e9 + 5;
;
const long double pi = 3.141592653589793238462643383279502884;
void input() {
  if (fopen("abc"
            ".inp",
            "r")) {
    freopen(
        "abc"
        ".inp",
        "r", stdin);
    freopen(
        "abc"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n;
int h[maxn], a[maxn];
long long s[maxn];
long long ans = 0, cur = 0;
vector<int> g[maxn];
void dfs(int u, int p) {
  cur += 1ll * h[u] * a[u];
  s[u] = a[u];
  for (int v : g[u]) {
    if (v != p) {
      h[v] = h[u] + 1;
      dfs(v, u);
      s[u] += s[v];
    }
  }
}
void re_root(int u, int p) {
  ans = max(ans, cur);
  for (int v : g[u]) {
    if (v == p) continue;
    cur -= s[v];
    s[u] -= s[v];
    cur += s[u];
    s[v] += s[u];
    re_root(v, u);
    s[v] -= s[u];
    cur -= s[u];
    s[u] += s[v];
    cur += s[v];
  }
}
int main() {
  input();
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  re_root(1, 0);
  cout << ans;
  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 maxn = 3e5 + 3;
const int maxm = 1e6 + 3;
const int mod = 1e9 + 7;
const int oo = 1e9 + 5;
;
const long double pi = 3.141592653589793238462643383279502884;
void input() {
  if (fopen("abc"
            ".inp",
            "r")) {
    freopen(
        "abc"
        ".inp",
        "r", stdin);
    freopen(
        "abc"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n;
int h[maxn], a[maxn];
long long s[maxn];
long long ans = 0, cur = 0;
vector<int> g[maxn];
void dfs(int u, int p) {
  cur += 1ll * h[u] * a[u];
  s[u] = a[u];
  for (int v : g[u]) {
    if (v != p) {
      h[v] = h[u] + 1;
      dfs(v, u);
      s[u] += s[v];
    }
  }
}
void re_root(int u, int p) {
  ans = max(ans, cur);
  for (int v : g[u]) {
    if (v == p) continue;
    cur -= s[v];
    s[u] -= s[v];
    cur += s[u];
    s[v] += s[u];
    re_root(v, u);
    s[v] -= s[u];
    cur -= s[u];
    s[u] += s[v];
    cur += s[v];
  }
}
int main() {
  input();
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  re_root(1, 0);
  cout << ans;
  return 0;
}

```