#include <bits/stdc++.h>
using namespace std;
const int maxn = 200200;
int N;
long long a[maxn];
long long zb[maxn];
long long curr[maxn];
long long rj[maxn];
long long sol;
int u;
int v;
vector<int> graf[maxn];
vector<int> stablo[maxn];
bool bio[maxn];
void make_tree(int x) {
  bio[x] = 1;
  for (int i = 0; i < graf[x].size(); i++) {
    int n = graf[x][i];
    if (!bio[n]) {
      stablo[x].push_back(n);
      make_tree(n);
    }
  }
  return;
}
void dfs(int x) {
  zb[x] = a[x];
  curr[x] = 0;
  for (int i = 0; i < stablo[x].size(); i++) {
    int n = stablo[x][i];
    dfs(n);
    zb[x] += zb[n];
    curr[x] += (curr[n] + zb[n]);
  }
  return;
}
void solve(int x, int p) {
  if (p) {
    rj[x] = rj[p] + zb[1] - 2 * zb[x];
    sol = max(sol, rj[x]);
  }
  for (int i = 0; i < stablo[x].size(); i++) solve(stablo[x][i], x);
  return;
}
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 = 1; i < N; i++) {
    cin >> u >> v;
    graf[u].push_back(v);
    graf[v].push_back(u);
  }
  make_tree(1);
  dfs(1);
  rj[1] = curr[1];
  sol = rj[1];
  solve(1, 0);
  cout << sol;
}

### 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;
const int maxn = 200200;
int N;
long long a[maxn];
long long zb[maxn];
long long curr[maxn];
long long rj[maxn];
long long sol;
int u;
int v;
vector<int> graf[maxn];
vector<int> stablo[maxn];
bool bio[maxn];
void make_tree(int x) {
  bio[x] = 1;
  for (int i = 0; i < graf[x].size(); i++) {
    int n = graf[x][i];
    if (!bio[n]) {
      stablo[x].push_back(n);
      make_tree(n);
    }
  }
  return;
}
void dfs(int x) {
  zb[x] = a[x];
  curr[x] = 0;
  for (int i = 0; i < stablo[x].size(); i++) {
    int n = stablo[x][i];
    dfs(n);
    zb[x] += zb[n];
    curr[x] += (curr[n] + zb[n]);
  }
  return;
}
void solve(int x, int p) {
  if (p) {
    rj[x] = rj[p] + zb[1] - 2 * zb[x];
    sol = max(sol, rj[x]);
  }
  for (int i = 0; i < stablo[x].size(); i++) solve(stablo[x][i], x);
  return;
}
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 = 1; i < N; i++) {
    cin >> u >> v;
    graf[u].push_back(v);
    graf[v].push_back(u);
  }
  make_tree(1);
  dfs(1);
  rj[1] = curr[1];
  sol = rj[1];
  solve(1, 0);
  cout << sol;
}

```