#include <bits/stdc++.h>
using namespace std;
long long int fun(vector<int> v[], long long int a[], long long int b[],
                  long long int d[], int ver, int par) {
  d[ver] = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) d[ver] = d[ver] + fun(v, a, b, d, v[ver][i], ver);
  }
  return d[ver] + b[ver] + a[ver];
}
long long int fun1(vector<int> v[], long long int a[], long long int b[],
                   int ver, int par) {
  b[ver] = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) b[ver] = b[ver] + fun1(v, a, b, v[ver][i], ver);
  }
  return b[ver] + a[ver];
}
void dfs(vector<int> v[], long long int a[], long long int b[],
         long long int c[], long long int d[], long long int x, int ver,
         int par, long long int ans[]) {
  long long int y;
  ans[ver] = d[ver] + x;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) {
      y = x + d[ver] - d[(v[ver][i])] - b[(v[ver][i])] - (2 * a[(v[ver][i])]) +
          c[(v[ver][i])];
      dfs(v, a, b, c, d, y, (v[ver][i]), ver, ans);
    }
  }
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  long long int a[n + 1], b[n + 1], c[n + 1], d[n + 1], ans[n + 1], sum = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  vector<int> v[n + 1];
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  b[0] = fun1(v, a, b, 1, 0);
  for (int i = 1; i < n + 1; i++) c[i] = sum - b[i];
  d[0] = fun(v, a, b, d, 1, 0);
  dfs(v, a, b, c, d, 0ll, 1, 0, ans);
  x = 1;
  for (int i = 1; i < n + 1; i++) {
    if (ans[x] < ans[i]) x = i;
  }
  printf("%lld ", ans[x]);
  printf("\n");
}

### 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;
long long int fun(vector<int> v[], long long int a[], long long int b[],
                  long long int d[], int ver, int par) {
  d[ver] = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) d[ver] = d[ver] + fun(v, a, b, d, v[ver][i], ver);
  }
  return d[ver] + b[ver] + a[ver];
}
long long int fun1(vector<int> v[], long long int a[], long long int b[],
                   int ver, int par) {
  b[ver] = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) b[ver] = b[ver] + fun1(v, a, b, v[ver][i], ver);
  }
  return b[ver] + a[ver];
}
void dfs(vector<int> v[], long long int a[], long long int b[],
         long long int c[], long long int d[], long long int x, int ver,
         int par, long long int ans[]) {
  long long int y;
  ans[ver] = d[ver] + x;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] != par) {
      y = x + d[ver] - d[(v[ver][i])] - b[(v[ver][i])] - (2 * a[(v[ver][i])]) +
          c[(v[ver][i])];
      dfs(v, a, b, c, d, y, (v[ver][i]), ver, ans);
    }
  }
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  long long int a[n + 1], b[n + 1], c[n + 1], d[n + 1], ans[n + 1], sum = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  vector<int> v[n + 1];
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  b[0] = fun1(v, a, b, 1, 0);
  for (int i = 1; i < n + 1; i++) c[i] = sum - b[i];
  d[0] = fun(v, a, b, d, 1, 0);
  dfs(v, a, b, c, d, 0ll, 1, 0, ans);
  x = 1;
  for (int i = 1; i < n + 1; i++) {
    if (ans[x] < ans[i]) x = i;
  }
  printf("%lld ", ans[x]);
  printf("\n");
}

```