#include <bits/stdc++.h>
using namespace std;
struct node {
  int64_t sm, val;
  node() { sm = val = 0; }
  node(const int64_t _sm, const int64_t _val) {
    sm = _sm;
    val = _val;
  }
  void operator+=(const node &other) {
    sm += other.sm;
    val += other.sm + other.val;
  }
};
int n;
int64_t as[200005];
vector<int> adj[200005];
node inf[200005];
node dfs(int st, int pr) {
  for (auto x : adj[st]) {
    if (x == pr) continue;
    inf[st] += dfs(x, st);
  }
  inf[st].sm += as[st];
  return inf[st];
}
int64_t dfs2(int st, int pr) {
  int64_t ans = inf[st].val;
  for (auto x : adj[st]) {
    if (x == pr) continue;
    inf[x].val = inf[st].val + inf[st].sm - inf[x].sm - inf[x].sm;
    inf[x].sm = inf[st].sm;
    ans = max(ans, dfs2(x, st));
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &as[i]);
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", dfs2(1, 0));
  return 0;
}

### Prompt

Please create a solution in cpp 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;
struct node {
  int64_t sm, val;
  node() { sm = val = 0; }
  node(const int64_t _sm, const int64_t _val) {
    sm = _sm;
    val = _val;
  }
  void operator+=(const node &other) {
    sm += other.sm;
    val += other.sm + other.val;
  }
};
int n;
int64_t as[200005];
vector<int> adj[200005];
node inf[200005];
node dfs(int st, int pr) {
  for (auto x : adj[st]) {
    if (x == pr) continue;
    inf[st] += dfs(x, st);
  }
  inf[st].sm += as[st];
  return inf[st];
}
int64_t dfs2(int st, int pr) {
  int64_t ans = inf[st].val;
  for (auto x : adj[st]) {
    if (x == pr) continue;
    inf[x].val = inf[st].val + inf[st].sm - inf[x].sm - inf[x].sm;
    inf[x].sm = inf[st].sm;
    ans = max(ans, dfs2(x, st));
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &as[i]);
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", dfs2(1, 0));
  return 0;
}

```