#include <bits/stdc++.h>
using namespace std;
int gcd(int f, int s) {
  if (s == 0)
    return f;
  else
    return gcd(s, f % s);
}
int const N = 1002006;
long long const M = 998244353;
long double const ep = .000000000000000001;
int n;
long long arr[N];
map<pair<int, int>, pair<long long, long long> > dp;
vector<int> G[N];
long long res[N];
pair<long long, long long> dfs(int node, int par) {
  pair<long long, long long> re = make_pair(0, 0);
  for (int i = 0; i < G[node].size(); i++) {
    int el = G[node][i];
    if (el != par) {
      pair<long long, long long> ok = dfs(el, node);
      re.first += ok.first;
      re.second += ok.second;
      res[node] += ok.first;
    }
  }
  dp[make_pair(node, par)] =
      make_pair(re.first + arr[node] + re.second, re.second + arr[node]);
  return dp[make_pair(node, par)];
}
void dfs1(int node, int par, long long curr, long long give) {
  res[node] += curr;
  for (int i = 0; i < G[node].size(); i++) {
    int el = G[node][i];
    if (el != par) {
      pair<long long, long long> ok = make_pair(
          dp[make_pair(node, par)].first - dp[make_pair(el, node)].first -
              dp[make_pair(node, par)].second,
          dp[make_pair(node, par)].second - dp[make_pair(el, node)].second -
              arr[node]);
      dfs1(el, node, ok.first + arr[node] + ok.second + curr + give,
           ok.second + arr[node] + give);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  for (int i = 0; i < n - 1; i++) {
    int temp, temp1;
    scanf("%d%d", &temp, &temp1);
    G[temp].push_back(temp1);
    G[temp1].push_back(temp);
  }
  dfs(1, -1);
  dfs1(1, -1, 0, 0);
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    mx = max(res[i], mx);
  }
  printf("%lld\n", mx);
  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>
using namespace std;
int gcd(int f, int s) {
  if (s == 0)
    return f;
  else
    return gcd(s, f % s);
}
int const N = 1002006;
long long const M = 998244353;
long double const ep = .000000000000000001;
int n;
long long arr[N];
map<pair<int, int>, pair<long long, long long> > dp;
vector<int> G[N];
long long res[N];
pair<long long, long long> dfs(int node, int par) {
  pair<long long, long long> re = make_pair(0, 0);
  for (int i = 0; i < G[node].size(); i++) {
    int el = G[node][i];
    if (el != par) {
      pair<long long, long long> ok = dfs(el, node);
      re.first += ok.first;
      re.second += ok.second;
      res[node] += ok.first;
    }
  }
  dp[make_pair(node, par)] =
      make_pair(re.first + arr[node] + re.second, re.second + arr[node]);
  return dp[make_pair(node, par)];
}
void dfs1(int node, int par, long long curr, long long give) {
  res[node] += curr;
  for (int i = 0; i < G[node].size(); i++) {
    int el = G[node][i];
    if (el != par) {
      pair<long long, long long> ok = make_pair(
          dp[make_pair(node, par)].first - dp[make_pair(el, node)].first -
              dp[make_pair(node, par)].second,
          dp[make_pair(node, par)].second - dp[make_pair(el, node)].second -
              arr[node]);
      dfs1(el, node, ok.first + arr[node] + ok.second + curr + give,
           ok.second + arr[node] + give);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  for (int i = 0; i < n - 1; i++) {
    int temp, temp1;
    scanf("%d%d", &temp, &temp1);
    G[temp].push_back(temp1);
    G[temp1].push_back(temp);
  }
  dfs(1, -1);
  dfs1(1, -1, 0, 0);
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    mx = max(res[i], mx);
  }
  printf("%lld\n", mx);
  return 0;
}

```