#include <bits/stdc++.h>
long long res, ans;
std::vector<long long> a, sum;
std::vector<std::vector<int>> g;
void dfs(int v, int p = -1, int h = 0) {
  res += 1LL * h * a[v];
  sum[v] = a[v];
  for (size_t i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, v, h + 1);
    sum[v] += sum[to];
  }
}
void dfs2(int v, int p = -1) {
  ans = std::max(ans, res);
  for (size_t i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    res -= sum[to];
    sum[v] -= sum[to];
    res += sum[v];
    sum[to] += sum[v];
    dfs2(to, v);
    sum[to] -= sum[v];
    res -= sum[v];
    sum[v] += sum[to];
    res += sum[to];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  a.resize(n);
  g.resize(n);
  sum.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a - 1].push_back(b - 1);
    g[b - 1].push_back(a - 1);
  }
  dfs(0);
  dfs2(0);
  printf("%lld\n", ans);
  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>
long long res, ans;
std::vector<long long> a, sum;
std::vector<std::vector<int>> g;
void dfs(int v, int p = -1, int h = 0) {
  res += 1LL * h * a[v];
  sum[v] = a[v];
  for (size_t i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, v, h + 1);
    sum[v] += sum[to];
  }
}
void dfs2(int v, int p = -1) {
  ans = std::max(ans, res);
  for (size_t i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    res -= sum[to];
    sum[v] -= sum[to];
    res += sum[v];
    sum[to] += sum[v];
    dfs2(to, v);
    sum[to] -= sum[v];
    res -= sum[v];
    sum[v] += sum[to];
    res += sum[to];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  a.resize(n);
  g.resize(n);
  sum.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a - 1].push_back(b - 1);
    g[b - 1].push_back(a - 1);
  }
  dfs(0);
  dfs2(0);
  printf("%lld\n", ans);
  return 0;
}

```