#include <bits/stdc++.h>
using namespace std;
int n;
long long a[200001];
long long son[200001], ans[200001];
vector<int> g[200001];
int dis[200001];
long long sum = 0;
void dfs(int cur, int d) {
  dis[cur] = d;
  son[cur] = a[cur];
  for (const auto nxt : g[cur]) {
    if (dis[nxt] < 0) {
      dfs(nxt, d + 1);
      son[cur] += son[nxt];
    }
  }
}
void calcu(int cur) {
  for (const auto &nxt : g[cur]) {
    if (!dis[nxt]) {
      dis[nxt] = 1;
      ans[nxt] = ans[cur] + sum - 2 * son[nxt];
      calcu(nxt);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dis, -1, sizeof(dis));
  cin >> n;
  int u, v;
  for (long long i = (1); i < (n + 1); ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long i = (0); i < (n - 1); ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  ans[1] = 0;
  for (long long i = (1); i < (n + 1); ++i) ans[1] += a[i] * dis[i];
  memset(dis, 0, sizeof(dis));
  dis[1] = 1;
  calcu(1);
  long long as = 0;
  for (long long i = (1); i < (n + 1); ++i) as = max(as, ans[i]);
  cout << as << 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>
using namespace std;
int n;
long long a[200001];
long long son[200001], ans[200001];
vector<int> g[200001];
int dis[200001];
long long sum = 0;
void dfs(int cur, int d) {
  dis[cur] = d;
  son[cur] = a[cur];
  for (const auto nxt : g[cur]) {
    if (dis[nxt] < 0) {
      dfs(nxt, d + 1);
      son[cur] += son[nxt];
    }
  }
}
void calcu(int cur) {
  for (const auto &nxt : g[cur]) {
    if (!dis[nxt]) {
      dis[nxt] = 1;
      ans[nxt] = ans[cur] + sum - 2 * son[nxt];
      calcu(nxt);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dis, -1, sizeof(dis));
  cin >> n;
  int u, v;
  for (long long i = (1); i < (n + 1); ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long i = (0); i < (n - 1); ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  ans[1] = 0;
  for (long long i = (1); i < (n + 1); ++i) ans[1] += a[i] * dis[i];
  memset(dis, 0, sizeof(dis));
  dis[1] = 1;
  calcu(1);
  long long as = 0;
  for (long long i = (1); i < (n + 1); ++i) as = max(as, ans[i]);
  cout << as << endl;
  return 0;
}

```