#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int d[500005], a[500005], c[500005];
vector<long long int> adj[500005];
long long int ans;
void dfs(long long int u, long long int par) {
  d[u] = a[u];
  for (long long int i = 0; i < adj[u].size(); i++) {
    long long int p = adj[u][i];
    if (p == par) continue;
    dfs(p, u);
    c[u] += c[p] + d[p];
    d[u] += d[p];
  }
}
void dfs1(long long int u, long long int par, long long int x,
          long long int y) {
  ans = max(ans, x);
  for (long long int i = 0; i < adj[u].size(); i++) {
    long long int p = adj[u][i];
    if (p == par) continue;
    long long int r = x - c[p] - d[p];
    long long int r1 = y - d[p];
    dfs1(p, u, c[p] + r + r1, d[p] + r1);
  }
}
int main() {
  long long int n, i, x, y;
  while (cin >> n) {
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i < n; i++) {
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1, -1);
    ans = 0;
    dfs1(1, -1, c[1], d[1]);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int d[500005], a[500005], c[500005];
vector<long long int> adj[500005];
long long int ans;
void dfs(long long int u, long long int par) {
  d[u] = a[u];
  for (long long int i = 0; i < adj[u].size(); i++) {
    long long int p = adj[u][i];
    if (p == par) continue;
    dfs(p, u);
    c[u] += c[p] + d[p];
    d[u] += d[p];
  }
}
void dfs1(long long int u, long long int par, long long int x,
          long long int y) {
  ans = max(ans, x);
  for (long long int i = 0; i < adj[u].size(); i++) {
    long long int p = adj[u][i];
    if (p == par) continue;
    long long int r = x - c[p] - d[p];
    long long int r1 = y - d[p];
    dfs1(p, u, c[p] + r + r1, d[p] + r1);
  }
}
int main() {
  long long int n, i, x, y;
  while (cin >> n) {
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i < n; i++) {
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1, -1);
    ans = 0;
    dfs1(1, -1, c[1], d[1]);
    cout << ans << endl;
  }
  return 0;
}

```