#include <bits/stdc++.h>
std::vector<int> e[200005];
int a[200005], n;
long long sum[200005], ans, cur, tot;
void dfs(int x, int p, int l) {
  sum[x] += a[x];
  cur += 1ll * l * a[x];
  for (int i = 0; i < e[x].size(); ++i) {
    if (e[x][i] == p) continue;
    dfs(e[x][i], x, l + 1);
    sum[x] += sum[e[x][i]];
  }
}
void dfs2(int x, int p) {
  ans = std::max(ans, cur);
  for (int i = 0; i < e[x].size(); ++i) {
    if (e[x][i] == p) continue;
    cur += tot - 2 * sum[e[x][i]];
    dfs2(e[x][i], x);
    cur -= tot - 2 * sum[e[x][i]];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    tot += a[i];
  }
  for (int i = 0, u, v; i < n - 1; ++i) {
    scanf("%d %d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, -1, 0);
  tot = sum[1];
  dfs2(1, -1);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
std::vector<int> e[200005];
int a[200005], n;
long long sum[200005], ans, cur, tot;
void dfs(int x, int p, int l) {
  sum[x] += a[x];
  cur += 1ll * l * a[x];
  for (int i = 0; i < e[x].size(); ++i) {
    if (e[x][i] == p) continue;
    dfs(e[x][i], x, l + 1);
    sum[x] += sum[e[x][i]];
  }
}
void dfs2(int x, int p) {
  ans = std::max(ans, cur);
  for (int i = 0; i < e[x].size(); ++i) {
    if (e[x][i] == p) continue;
    cur += tot - 2 * sum[e[x][i]];
    dfs2(e[x][i], x);
    cur -= tot - 2 * sum[e[x][i]];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    tot += a[i];
  }
  for (int i = 0, u, v; i < n - 1; ++i) {
    scanf("%d %d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, -1, 0);
  tot = sum[1];
  dfs2(1, -1);
  printf("%I64d\n", ans);
  return 0;
}

```