#include <bits/stdc++.h>
using namespace std;
template <typename T, typename V>
void bugp(const pair<T, V> &x) {
  cerr << '{' << x.first << ", " << x.second << '}' << endl;
}
template <typename T, typename U, typename V>
void bugpp(const pair<T, pair<U, V> > &x) {
  cerr << '{' << x.first << ", {" << x.second.first << ", " << x.second.second
       << "}}" << endl;
}
template <typename T>
bool maximize(T &x, const T &y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T>
bool minimize(T &x, const T &y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
const int N = 200002;
int n, wnode[N], H[N];
long long sumW[N], res[N], total = 0;
vector<int> G[N];
void calc(int u, int par) {
  res[1] += 1LL * H[u] * wnode[u];
  sumW[u] = wnode[u];
  for (const int &v : G[u]) {
    if (v == par) continue;
    H[v] = H[u] + 1;
    calc(v, u);
    sumW[u] += sumW[v];
  }
}
void DFS(int u, int par) {
  for (const int &v : G[u]) {
    if (v == par) continue;
    res[v] = res[u] + total - 2 * sumW[v];
    DFS(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1, _n = (n); i <= _n; ++i) {
    cin >> wnode[i];
    total += wnode[i];
  }
  for (int i = 1, _n = (n - 1); i <= _n; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  calc(1, 1);
  DFS(1, 1);
  cout << *max_element(res + 1, res + n + 1);
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T, typename V>
void bugp(const pair<T, V> &x) {
  cerr << '{' << x.first << ", " << x.second << '}' << endl;
}
template <typename T, typename U, typename V>
void bugpp(const pair<T, pair<U, V> > &x) {
  cerr << '{' << x.first << ", {" << x.second.first << ", " << x.second.second
       << "}}" << endl;
}
template <typename T>
bool maximize(T &x, const T &y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T>
bool minimize(T &x, const T &y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
const int N = 200002;
int n, wnode[N], H[N];
long long sumW[N], res[N], total = 0;
vector<int> G[N];
void calc(int u, int par) {
  res[1] += 1LL * H[u] * wnode[u];
  sumW[u] = wnode[u];
  for (const int &v : G[u]) {
    if (v == par) continue;
    H[v] = H[u] + 1;
    calc(v, u);
    sumW[u] += sumW[v];
  }
}
void DFS(int u, int par) {
  for (const int &v : G[u]) {
    if (v == par) continue;
    res[v] = res[u] + total - 2 * sumW[v];
    DFS(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1, _n = (n); i <= _n; ++i) {
    cin >> wnode[i];
    total += wnode[i];
  }
  for (int i = 1, _n = (n - 1); i <= _n; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  calc(1, 1);
  DFS(1, 1);
  cout << *max_element(res + 1, res + n + 1);
  return 0;
}

```