#include <bits/stdc++.h>
using namespace std;
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
const long long MOD = 1e9 + 7;
const long long INF = 1e14;
long long mpow(long long a, long long b, long long p = MOD) {
  a = a % p;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b = b >> 1;
  }
  return res % p;
}
const long long N = 1000002;
vector<long long> a[N];
long long dis[N];
long long s[N];
long long ans[N];
long long an[N];
void dfs(long long u, long long p = -1, long long d = 0) {
  dis[u] = d;
  s[u] = an[u];
  for (auto x : a[u]) {
    if (x != p) {
      dfs(x, u, d + 1);
      s[u] += s[x];
    }
  }
}
void dfs1(long long u, long long p = -1) {
  for (auto x : a[u]) {
    if (x != p) {
      ans[x] = ans[u] - s[x] + s[1] - s[x];
      dfs1(x, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (int i = 1; i <= (int)(n); ++i) cin >> an[i];
  for (int i = 0; i < (int)(n - 1); ++i) {
    long long x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1);
  for (int i = 1; i <= (int)(n); ++i) ans[1] += dis[i] * an[i];
  dfs1(1);
  cout << *max_element(ans, ans + N);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
const long long MOD = 1e9 + 7;
const long long INF = 1e14;
long long mpow(long long a, long long b, long long p = MOD) {
  a = a % p;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b = b >> 1;
  }
  return res % p;
}
const long long N = 1000002;
vector<long long> a[N];
long long dis[N];
long long s[N];
long long ans[N];
long long an[N];
void dfs(long long u, long long p = -1, long long d = 0) {
  dis[u] = d;
  s[u] = an[u];
  for (auto x : a[u]) {
    if (x != p) {
      dfs(x, u, d + 1);
      s[u] += s[x];
    }
  }
}
void dfs1(long long u, long long p = -1) {
  for (auto x : a[u]) {
    if (x != p) {
      ans[x] = ans[u] - s[x] + s[1] - s[x];
      dfs1(x, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (int i = 1; i <= (int)(n); ++i) cin >> an[i];
  for (int i = 0; i < (int)(n - 1); ++i) {
    long long x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1);
  for (int i = 1; i <= (int)(n); ++i) ans[1] += dis[i] * an[i];
  dfs1(1);
  cout << *max_element(ans, ans + N);
}

```