#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
bool se[1000001] = {0};
void seive(long long int n) {
  for (long long int i = 2; i <= n; i++) {
    if (se[i]) {
      continue;
    }
    for (long long int u = 2 * i; u <= n; u += i) {
      se[u] = true;
    }
  }
}
vector<long long int> g[200001];
long long int sum[200001];
long long int a[200001];
long long int res = 0;
long long int ans = 0;
void dfs(long long int u, long long int pu, long long int h) {
  res += h * a[u];
  sum[u] = a[u];
  for (auto ad : g[u]) {
    if (ad != pu) {
      dfs(ad, u, h + 1);
      sum[u] += sum[ad];
    }
  }
}
void dfs2(long long int u, long long int pu) {
  ans = max(ans, res);
  for (auto ad : g[u]) {
    if (ad != pu) {
      res -= sum[ad];
      sum[u] -= sum[ad];
      res += sum[u];
      sum[ad] += sum[u];
      dfs2(ad, u);
      sum[ad] -= sum[u];
      res -= sum[u];
      sum[u] += sum[ad];
      res += sum[ad];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n - 1; i++) {
      long long int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1, 0, 0);
    dfs2(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;
long long int mod = 1000000007;
bool se[1000001] = {0};
void seive(long long int n) {
  for (long long int i = 2; i <= n; i++) {
    if (se[i]) {
      continue;
    }
    for (long long int u = 2 * i; u <= n; u += i) {
      se[u] = true;
    }
  }
}
vector<long long int> g[200001];
long long int sum[200001];
long long int a[200001];
long long int res = 0;
long long int ans = 0;
void dfs(long long int u, long long int pu, long long int h) {
  res += h * a[u];
  sum[u] = a[u];
  for (auto ad : g[u]) {
    if (ad != pu) {
      dfs(ad, u, h + 1);
      sum[u] += sum[ad];
    }
  }
}
void dfs2(long long int u, long long int pu) {
  ans = max(ans, res);
  for (auto ad : g[u]) {
    if (ad != pu) {
      res -= sum[ad];
      sum[u] -= sum[ad];
      res += sum[u];
      sum[ad] += sum[u];
      dfs2(ad, u);
      sum[ad] -= sum[u];
      res -= sum[u];
      sum[u] += sum[ad];
      res += sum[ad];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n - 1; i++) {
      long long int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1, 0, 0);
    dfs2(1, 0);
    cout << ans;
  }
  return 0;
}

```