#include <bits/stdc++.h>
using namespace std;
vector<long long> c;
vector<long long> st;
long long sum = 0, s = 0;
vector<vector<long long> > Gr;
struct Ver {
  Ver() {}
  Ver(long long a, long long b, long long c) {
    now = a;
    pre = b;
    dist = c;
  }
  long long now;
  long long pre;
  long long dist;
};
vector<long long> Ans;
void calc(Ver a) {
  s += a.dist * c[a.now];
  for (int i = 0; i < Gr[a.now].size(); i++) {
    if (Gr[a.now][i] != a.pre) {
      calc(Ver(Gr[a.now][i], a.now, a.dist + 1));
      st[a.now] += st[Gr[a.now][i]];
    }
  }
  st[a.now] += c[a.now];
}
void dp(Ver a) {
  for (int i = 0; i < Gr[a.now].size(); i++) {
    if (Gr[a.now][i] != a.pre) {
      Ans[Gr[a.now][i]] =
          Ans[a.now] - st[Gr[a.now][i]] + sum - st[Gr[a.now][i]];
      dp(Ver(Gr[a.now][i], a.now, -1));
    }
  }
}
int main() {
  long long N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    c.push_back(a);
    sum += a;
  }
  Gr.resize(N);
  st.resize(N, 0);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    Gr[a].push_back(b);
    Gr[b].push_back(a);
  }
  calc(Ver(0, -1, 0));
  Ans.resize(N, 0);
  Ans[0] = s;
  dp(Ver(0, -1, -1));
  long long r = 0;
  for (int i = 0; i < Ans.size(); i++) {
    r = max((long long)Ans[i], r);
  }
  cout << r;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<long long> c;
vector<long long> st;
long long sum = 0, s = 0;
vector<vector<long long> > Gr;
struct Ver {
  Ver() {}
  Ver(long long a, long long b, long long c) {
    now = a;
    pre = b;
    dist = c;
  }
  long long now;
  long long pre;
  long long dist;
};
vector<long long> Ans;
void calc(Ver a) {
  s += a.dist * c[a.now];
  for (int i = 0; i < Gr[a.now].size(); i++) {
    if (Gr[a.now][i] != a.pre) {
      calc(Ver(Gr[a.now][i], a.now, a.dist + 1));
      st[a.now] += st[Gr[a.now][i]];
    }
  }
  st[a.now] += c[a.now];
}
void dp(Ver a) {
  for (int i = 0; i < Gr[a.now].size(); i++) {
    if (Gr[a.now][i] != a.pre) {
      Ans[Gr[a.now][i]] =
          Ans[a.now] - st[Gr[a.now][i]] + sum - st[Gr[a.now][i]];
      dp(Ver(Gr[a.now][i], a.now, -1));
    }
  }
}
int main() {
  long long N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    c.push_back(a);
    sum += a;
  }
  Gr.resize(N);
  st.resize(N, 0);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    Gr[a].push_back(b);
    Gr[b].push_back(a);
  }
  calc(Ver(0, -1, 0));
  Ans.resize(N, 0);
  Ans[0] = s;
  dp(Ver(0, -1, -1));
  long long r = 0;
  for (int i = 0; i < Ans.size(); i++) {
    r = max((long long)Ans[i], r);
  }
  cout << r;
  return 0;
}

```