#include <bits/stdc++.h>
using namespace std;
vector<long long> node[200001];
long long sz[200001] = {}, vis[200001] = {}, a[200001] = {}, depth[200001] = {};
long long ans[200001] = {}, cnt = 0, ma = 0;
void dfs(long long s) {
  cnt += depth[s] * a[s];
  sz[s] = a[s];
  vis[s] = 1;
  for (long long i : node[s]) {
    if (!vis[i]) {
      depth[i] = depth[s] + 1;
      dfs(i);
      sz[s] += sz[i];
    }
  }
}
void dfs2(long long s) {
  vis[s] = 1;
  ma = max(ma, ans[s]);
  for (long long i : node[s]) {
    if (!vis[i]) {
      ans[i] = ans[s] + (sz[1] - sz[i]) - sz[i];
      dfs2(i);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    node[u].push_back(v);
    node[v].push_back(u);
  }
  dfs(1);
  ans[1] = cnt;
  for (long long i = 0; i <= 200000; i++) vis[i] = 0;
  dfs2(1);
  cout << ma;
}

### Prompt

Please create a solution in CPP 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;
vector<long long> node[200001];
long long sz[200001] = {}, vis[200001] = {}, a[200001] = {}, depth[200001] = {};
long long ans[200001] = {}, cnt = 0, ma = 0;
void dfs(long long s) {
  cnt += depth[s] * a[s];
  sz[s] = a[s];
  vis[s] = 1;
  for (long long i : node[s]) {
    if (!vis[i]) {
      depth[i] = depth[s] + 1;
      dfs(i);
      sz[s] += sz[i];
    }
  }
}
void dfs2(long long s) {
  vis[s] = 1;
  ma = max(ma, ans[s]);
  for (long long i : node[s]) {
    if (!vis[i]) {
      ans[i] = ans[s] + (sz[1] - sz[i]) - sz[i];
      dfs2(i);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    node[u].push_back(v);
    node[v].push_back(u);
  }
  dfs(1);
  ans[1] = cnt;
  for (long long i = 0; i <= 200000; i++) vis[i] = 0;
  dfs2(1);
  cout << ma;
}

```