#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
vector<int> adj[MAXN];
vector<int> visited(MAXN, 0);
long long int sub[MAXN];
long long int a[MAXN];
long long int res[MAXN];
int n;
long long int sum = 0;
void dfs(int s, int p) {
  for (auto u : adj[s]) {
    if (u != p) {
      dfs(u, s);
      sub[s] += sub[u];
    }
  }
}
vector<long long int> dist(MAXN, INT_MAX);
void bfs(int s) {
  dist[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < adj[v].size(); i++) {
      int u = adj[v][i];
      if (dist[u] == INT_MAX) {
        dist[u] = dist[v] + 1;
        q.push(u);
      }
    }
  }
}
vector<int> dist2(MAXN, INT_MAX);
void bfs2(int s) {
  dist2[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < adj[v].size(); i++) {
      int u = adj[v][i];
      if (dist2[u] == INT_MAX) {
        dist2[u] = dist2[v] + 1;
        q.push(u);
        res[u] = res[v] + sum - 2 * sub[u];
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sub[i] = a[i];
    sum += a[i];
  }
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  bfs(1);
  memset(res, 0, sizeof(res));
  for (int i = 1; i <= n; i++) res[1] += dist[i] * a[i];
  dfs(1, 0);
  bfs2(1);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) ans = max(ans, res[i]);
  cout << ans << "\n";
  return 0;
}

### 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 int MAXN = 200005;
vector<int> adj[MAXN];
vector<int> visited(MAXN, 0);
long long int sub[MAXN];
long long int a[MAXN];
long long int res[MAXN];
int n;
long long int sum = 0;
void dfs(int s, int p) {
  for (auto u : adj[s]) {
    if (u != p) {
      dfs(u, s);
      sub[s] += sub[u];
    }
  }
}
vector<long long int> dist(MAXN, INT_MAX);
void bfs(int s) {
  dist[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < adj[v].size(); i++) {
      int u = adj[v][i];
      if (dist[u] == INT_MAX) {
        dist[u] = dist[v] + 1;
        q.push(u);
      }
    }
  }
}
vector<int> dist2(MAXN, INT_MAX);
void bfs2(int s) {
  dist2[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < adj[v].size(); i++) {
      int u = adj[v][i];
      if (dist2[u] == INT_MAX) {
        dist2[u] = dist2[v] + 1;
        q.push(u);
        res[u] = res[v] + sum - 2 * sub[u];
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sub[i] = a[i];
    sum += a[i];
  }
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  bfs(1);
  memset(res, 0, sizeof(res));
  for (int i = 1; i <= n; i++) res[1] += dist[i] * a[i];
  dfs(1, 0);
  bfs2(1);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) ans = max(ans, res[i]);
  cout << ans << "\n";
  return 0;
}

```