#include <bits/stdc++.h>
using namespace std;
const char nxtl = '\n';
const int MAXN = (int)5e5 + 228, mod = (int)1e9 + 7;
const double eps = (double)1e-7;
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
int n, a[MAXN];
long long dp[MAXN], res;
vector<int> g[MAXN];
void dfs1(int v, int pr, int depth) {
  dp[v] = a[v];
  res += a[v] * 1ll * depth;
  for (auto &to : g[v]) {
    if (to == pr) continue;
    dfs1(to, v, depth + 1);
    dp[v] += dp[to];
  }
}
void dfs2(int v, int pr, long long cur) {
  updmax(res, cur);
  for (auto &to : g[v]) {
    if (to == pr) continue;
    dfs2(to, v, cur + dp[1] - (dp[to] << 1));
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned int FOR;
  asm("rdtsc" : "=A"(FOR));
  srand(FOR);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1(1, -1, 0);
  dfs2(1, -1, res);
  cout << res << nxtl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const char nxtl = '\n';
const int MAXN = (int)5e5 + 228, mod = (int)1e9 + 7;
const double eps = (double)1e-7;
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
int n, a[MAXN];
long long dp[MAXN], res;
vector<int> g[MAXN];
void dfs1(int v, int pr, int depth) {
  dp[v] = a[v];
  res += a[v] * 1ll * depth;
  for (auto &to : g[v]) {
    if (to == pr) continue;
    dfs1(to, v, depth + 1);
    dp[v] += dp[to];
  }
}
void dfs2(int v, int pr, long long cur) {
  updmax(res, cur);
  for (auto &to : g[v]) {
    if (to == pr) continue;
    dfs2(to, v, cur + dp[1] - (dp[to] << 1));
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned int FOR;
  asm("rdtsc" : "=A"(FOR));
  srand(FOR);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1(1, -1, 0);
  dfs2(1, -1, res);
  cout << res << nxtl;
  return 0;
}

```