#include <bits/stdc++.h>
using namespace std;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 2e5 + 7;
int n;
long long a[maxn];
set<int> g[maxn];
long long tot;
long long sub[maxn];
long long ans;
void dfs1(int u, int p, int d) {
  tot += 1LL * a[u] * d;
  sub[u] += a[u];
  for (int v : g[u]) {
    if (v == p) continue;
    dfs1(v, u, d + 1);
    sub[u] += sub[v];
  }
}
void dfs2(int at, int p) {
  ans = max(ans, tot);
  for (int to : g[at]) {
    if (to == p) continue;
    tot -= sub[to];
    sub[at] -= sub[to];
    tot += sub[at];
    sub[to] += sub[at];
    dfs2(to, at);
    sub[to] -= sub[at];
    tot -= sub[at];
    sub[at] += sub[to];
    tot += sub[to];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v);
    g[v].insert(u);
  }
  tot = 0LL;
  dfs1(1, -1, 0);
  ans = tot;
  dfs2(1, -1);
  out(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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 2e5 + 7;
int n;
long long a[maxn];
set<int> g[maxn];
long long tot;
long long sub[maxn];
long long ans;
void dfs1(int u, int p, int d) {
  tot += 1LL * a[u] * d;
  sub[u] += a[u];
  for (int v : g[u]) {
    if (v == p) continue;
    dfs1(v, u, d + 1);
    sub[u] += sub[v];
  }
}
void dfs2(int at, int p) {
  ans = max(ans, tot);
  for (int to : g[at]) {
    if (to == p) continue;
    tot -= sub[to];
    sub[at] -= sub[to];
    tot += sub[at];
    sub[to] += sub[at];
    dfs2(to, at);
    sub[to] -= sub[at];
    tot -= sub[at];
    sub[at] += sub[to];
    tot += sub[to];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v);
    g[v].insert(u);
  }
  tot = 0LL;
  dfs1(1, -1, 0);
  ans = tot;
  dfs2(1, -1);
  out(ans);
  return 0;
}

```