#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 200005;
vector<int> G[MAXN];
ll weight[MAXN], sum[MAXN], DP[MAXN], res;
int n, depth[MAXN];
void build_sum(int id, int fa) {
  sum[id] = weight[id];
  for (auto x : G[id]) {
    if (x != fa) {
      depth[x] = depth[id] + 1;
      build_sum(x, id);
      sum[id] += sum[x];
    }
  }
}
void dfs(int id, int fa) {
  res = max(res, DP[id]);
  for (auto x : G[id]) {
    if (x != fa) {
      DP[x] = DP[id] + sum[1] - 2 * sum[x];
      dfs(x, id);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int u, v;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> weight[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  depth[1] = 0;
  build_sum(1, 1);
  res = DP[1] = 0;
  for (int i = 2; i <= n; ++i) {
    DP[1] += weight[i] * depth[i];
  }
  dfs(1, 1);
  cout << res << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
const int MAXN = 200005;
vector<int> G[MAXN];
ll weight[MAXN], sum[MAXN], DP[MAXN], res;
int n, depth[MAXN];
void build_sum(int id, int fa) {
  sum[id] = weight[id];
  for (auto x : G[id]) {
    if (x != fa) {
      depth[x] = depth[id] + 1;
      build_sum(x, id);
      sum[id] += sum[x];
    }
  }
}
void dfs(int id, int fa) {
  res = max(res, DP[id]);
  for (auto x : G[id]) {
    if (x != fa) {
      DP[x] = DP[id] + sum[1] - 2 * sum[x];
      dfs(x, id);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int u, v;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> weight[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  depth[1] = 0;
  build_sum(1, 1);
  res = DP[1] = 0;
  for (int i = 2; i <= n; ++i) {
    DP[1] += weight[i] * depth[i];
  }
  dfs(1, 1);
  cout << res << '\n';
  return 0;
}

```