#include <bits/stdc++.h>
using namespace std;
inline void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long int mod = (long long int)(1e9) + 7;
long long int pmod(long long int x, long long int n) {
  long long int r = 1;
  while (n > 0) {
    if (n & 1LL) r = (r * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return r % mod;
}
long long int n;
const int mxn = (int)(2e5);
vector<int> adj[mxn + 1];
long long int a[mxn + 1];
long long int dp1[mxn + 1], dp2[mxn + 1], dp3[mxn + 1];
long long int sb[mxn + 1];
void dfs(int u, int p) {
  sb[u] = a[u];
  for (auto it : adj[u]) {
    if (it == p) continue;
    dfs(it, u);
    sb[u] += sb[it];
    dp1[u] += (dp1[it] + sb[it]);
  }
}
void dfs1(int u, int p) {
  dp3[u] = dp1[u] + dp2[u];
  for (auto it : adj[u]) {
    if (it == p) continue;
    dp2[it] = dp2[u] + (dp1[u] - dp1[it] - sb[it]) + (sb[1] - sb[it]);
    dfs1(it, u);
  }
}
void solve() {
  cin >> n;
  int x, y;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  long long int mx = 0;
  for (int i = 1; i <= n; i++) {
    mx = max(mx, dp3[i]);
  }
  cout << mx;
}
int main() {
  fast();
  solve();
}

### 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;
inline void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long int mod = (long long int)(1e9) + 7;
long long int pmod(long long int x, long long int n) {
  long long int r = 1;
  while (n > 0) {
    if (n & 1LL) r = (r * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return r % mod;
}
long long int n;
const int mxn = (int)(2e5);
vector<int> adj[mxn + 1];
long long int a[mxn + 1];
long long int dp1[mxn + 1], dp2[mxn + 1], dp3[mxn + 1];
long long int sb[mxn + 1];
void dfs(int u, int p) {
  sb[u] = a[u];
  for (auto it : adj[u]) {
    if (it == p) continue;
    dfs(it, u);
    sb[u] += sb[it];
    dp1[u] += (dp1[it] + sb[it]);
  }
}
void dfs1(int u, int p) {
  dp3[u] = dp1[u] + dp2[u];
  for (auto it : adj[u]) {
    if (it == p) continue;
    dp2[it] = dp2[u] + (dp1[u] - dp1[it] - sb[it]) + (sb[1] - sb[it]);
    dfs1(it, u);
  }
}
void solve() {
  cin >> n;
  int x, y;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  long long int mx = 0;
  for (int i = 1; i <= n; i++) {
    mx = max(mx, dp3[i]);
  }
  cout << mx;
}
int main() {
  fast();
  solve();
}

```