#include <bits/stdc++.h>
using namespace std;
const long double PI = 2.0 * acos(0);
const long long M = 1000000007;
long long binpow(long long a, long long b) {
  if (!b) return 1ll;
  long long temp = binpow(a, b >> 1ll);
  temp = (1ll * temp * temp) % M;
  if (b & 1) temp = (1ll * a * temp) % M;
  return temp;
}
const long long N = 200100;
std::vector<long long> adj[N];
long long a[N], ans[N], parent[N], par[N], sum[N], totsum;
bool vis[N];
long long n;
void dfs(long long u) {
  vis[u] = 1;
  sum[u] = a[u];
  for (long long &v : adj[u]) {
    if (!vis[v]) {
      dfs(v);
      ans[u] += (ans[v] + sum[v]);
      sum[u] += sum[v];
      par[v] = u;
    }
  }
}
void dfs2(long long u) {
  parent[u] = parent[par[u]] + totsum - sum[par[u]];
  parent[u] += (ans[par[u]] - ans[u] - sum[u]);
  parent[u] += (sum[par[u]] - sum[u]);
  for (long long v : adj[u])
    if (v != par[u]) dfs2(v);
}
void solve() {
  long long m, k, i, j, l, r;
  bool flag;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i + 1];
    totsum += a[i + 1];
  }
  for (long long i = 0; i < n - 1; ++i) {
    cin >> j >> k;
    adj[j].emplace_back(k);
    adj[k].emplace_back(j);
  }
  dfs(1);
  for (long long v : adj[1]) dfs2(v);
  long long res = 0;
  for (long long i = 0; i < n; ++i) res = max(res, ans[i + 1] + parent[i + 1]);
  cout << res << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  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;
const long double PI = 2.0 * acos(0);
const long long M = 1000000007;
long long binpow(long long a, long long b) {
  if (!b) return 1ll;
  long long temp = binpow(a, b >> 1ll);
  temp = (1ll * temp * temp) % M;
  if (b & 1) temp = (1ll * a * temp) % M;
  return temp;
}
const long long N = 200100;
std::vector<long long> adj[N];
long long a[N], ans[N], parent[N], par[N], sum[N], totsum;
bool vis[N];
long long n;
void dfs(long long u) {
  vis[u] = 1;
  sum[u] = a[u];
  for (long long &v : adj[u]) {
    if (!vis[v]) {
      dfs(v);
      ans[u] += (ans[v] + sum[v]);
      sum[u] += sum[v];
      par[v] = u;
    }
  }
}
void dfs2(long long u) {
  parent[u] = parent[par[u]] + totsum - sum[par[u]];
  parent[u] += (ans[par[u]] - ans[u] - sum[u]);
  parent[u] += (sum[par[u]] - sum[u]);
  for (long long v : adj[u])
    if (v != par[u]) dfs2(v);
}
void solve() {
  long long m, k, i, j, l, r;
  bool flag;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i + 1];
    totsum += a[i + 1];
  }
  for (long long i = 0; i < n - 1; ++i) {
    cin >> j >> k;
    adj[j].emplace_back(k);
    adj[k].emplace_back(j);
  }
  dfs(1);
  for (long long v : adj[1]) dfs2(v);
  long long res = 0;
  for (long long i = 0; i < n; ++i) res = max(res, ans[i + 1] + parent[i + 1]);
  cout << res << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```