#include <bits/stdc++.h>
using namespace std;
struct side {
  int to, next;
} a[400001];
long long f[200001], siz[200001], ans;
int s[200001], n, root, cnt;
inline void mem(int q, int w) {
  a[++cnt].to = w;
  a[cnt].next = s[q];
  s[q] = cnt;
  return;
}
inline void dfs(int x, int fa, long long len) {
  f[root] += len * siz[x];
  int w = s[x];
  while (w) {
    if (a[w].to != fa) {
      dfs(a[w].to, x, len + 1);
      siz[x] += siz[a[w].to];
    }
    w = a[w].next;
  }
  return;
}
inline void dp(int x, int fa) {
  int w = s[x];
  while (w) {
    if (a[w].to != fa) {
      f[a[w].to] = f[x] + siz[root] - 2 * siz[a[w].to];
      dp(a[w].to, x);
    }
    w = a[w].next;
  }
  if (ans < f[x]) ans = f[x];
  return;
}
int main() {
  std::ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> siz[i];
  for (int i = 1; i < n; i++) {
    int q, w;
    cin >> q >> w;
    mem(q, w);
    mem(w, q);
  }
  srand(time(0));
  root = rand() % n + 1;
  dfs(root, 0, 0);
  dp(root, 0);
  cout << 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>
using namespace std;
struct side {
  int to, next;
} a[400001];
long long f[200001], siz[200001], ans;
int s[200001], n, root, cnt;
inline void mem(int q, int w) {
  a[++cnt].to = w;
  a[cnt].next = s[q];
  s[q] = cnt;
  return;
}
inline void dfs(int x, int fa, long long len) {
  f[root] += len * siz[x];
  int w = s[x];
  while (w) {
    if (a[w].to != fa) {
      dfs(a[w].to, x, len + 1);
      siz[x] += siz[a[w].to];
    }
    w = a[w].next;
  }
  return;
}
inline void dp(int x, int fa) {
  int w = s[x];
  while (w) {
    if (a[w].to != fa) {
      f[a[w].to] = f[x] + siz[root] - 2 * siz[a[w].to];
      dp(a[w].to, x);
    }
    w = a[w].next;
  }
  if (ans < f[x]) ans = f[x];
  return;
}
int main() {
  std::ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> siz[i];
  for (int i = 1; i < n; i++) {
    int q, w;
    cin >> q >> w;
    mem(q, w);
    mem(w, q);
  }
  srand(time(0));
  root = rand() % n + 1;
  dfs(root, 0, 0);
  dp(root, 0);
  cout << ans;
  return 0;
}

```