#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 1;
vector<long long> g[N];
long long cost[N], x, y, ans, sum, ssum[N], sure_ans, n;
bool visited[N], visited1[N];
void dfs(long long v, long long d) {
  ans += d * cost[v];
  ssum[v] = cost[v];
  visited[v] = 1;
  d++;
  for (long long i = 0; i < (long long)(g[v].size()); i++) {
    if (!visited[g[v][i]]) {
      dfs(g[v][i], d);
      ssum[v] += ssum[g[v][i]];
    }
  }
}
void max_ans(long long v, long long t_ans) {
  if (v == 1)
    sure_ans = t_ans;
  else {
    t_ans = t_ans + sum - 2 * ssum[v];
    sure_ans = max(sure_ans, t_ans);
  }
  visited1[v] = 1;
  for (long long i = 0; i < (long long)(g[v].size()); i++) {
    if (!visited1[g[v][i]]) {
      max_ans(g[v][i], t_ans);
    }
  }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> cost[i];
    sum += cost[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  max_ans(1, ans);
  cout << sure_ans << endl;
}

### Prompt

Please formulate 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 long long N = 2e5 + 1;
vector<long long> g[N];
long long cost[N], x, y, ans, sum, ssum[N], sure_ans, n;
bool visited[N], visited1[N];
void dfs(long long v, long long d) {
  ans += d * cost[v];
  ssum[v] = cost[v];
  visited[v] = 1;
  d++;
  for (long long i = 0; i < (long long)(g[v].size()); i++) {
    if (!visited[g[v][i]]) {
      dfs(g[v][i], d);
      ssum[v] += ssum[g[v][i]];
    }
  }
}
void max_ans(long long v, long long t_ans) {
  if (v == 1)
    sure_ans = t_ans;
  else {
    t_ans = t_ans + sum - 2 * ssum[v];
    sure_ans = max(sure_ans, t_ans);
  }
  visited1[v] = 1;
  for (long long i = 0; i < (long long)(g[v].size()); i++) {
    if (!visited1[g[v][i]]) {
      max_ans(g[v][i], t_ans);
    }
  }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> cost[i];
    sum += cost[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  max_ans(1, ans);
  cout << sure_ans << endl;
}

```