#include <bits/stdc++.h>
using namespace std;
const long long INF = 4 * 1e18;
const int INFi = 2 * 1e9;
long long mod = 1e9 + 7;
vector<long long> sz;
vector<long long> cnt;
vector<long long> a;
vector<vector<int>> edge;
void dfs_sz(int v, int p) {
  sz[v] = 0;
  for (int i : edge[v]) {
    if (i == p) continue;
    dfs_sz(i, v);
    sz[v] += sz[i] + a[i];
  }
}
void dfs_cnt(int v, int p) {
  for (int u : edge[v]) {
    if (u == p) continue;
    dfs_cnt(u, v);
    cnt[v] += cnt[u];
  }
  cnt[v] += sz[v];
}
long long maxi = 0;
void dfs(int v, int p, long long cur, long long tot) {
  maxi = max(maxi, tot + cnt[v]);
  for (int u : edge[v]) {
    if (u == p) continue;
    dfs(u, v, cur + sz[v] - sz[u] - a[u] + a[v],
        tot + cur + cnt[v] + sz[v] + a[v] - 2 * sz[u] - cnt[u] - 2 * a[u]);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  sz.resize(n);
  a.resize(n);
  edge.resize(n);
  cnt.resize(n);
  int t1, t2;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    cin >> t1 >> t2;
    t1--;
    t2--;
    edge[t2].push_back(t1);
    edge[t1].push_back(t2);
  }
  dfs_sz(0, -1);
  dfs_cnt(0, -1);
  for (int i = 0; i < n; i++) {
  }
  dfs(0, -1, 0, 0);
  cout << maxi << endl;
  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 long long INF = 4 * 1e18;
const int INFi = 2 * 1e9;
long long mod = 1e9 + 7;
vector<long long> sz;
vector<long long> cnt;
vector<long long> a;
vector<vector<int>> edge;
void dfs_sz(int v, int p) {
  sz[v] = 0;
  for (int i : edge[v]) {
    if (i == p) continue;
    dfs_sz(i, v);
    sz[v] += sz[i] + a[i];
  }
}
void dfs_cnt(int v, int p) {
  for (int u : edge[v]) {
    if (u == p) continue;
    dfs_cnt(u, v);
    cnt[v] += cnt[u];
  }
  cnt[v] += sz[v];
}
long long maxi = 0;
void dfs(int v, int p, long long cur, long long tot) {
  maxi = max(maxi, tot + cnt[v]);
  for (int u : edge[v]) {
    if (u == p) continue;
    dfs(u, v, cur + sz[v] - sz[u] - a[u] + a[v],
        tot + cur + cnt[v] + sz[v] + a[v] - 2 * sz[u] - cnt[u] - 2 * a[u]);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  sz.resize(n);
  a.resize(n);
  edge.resize(n);
  cnt.resize(n);
  int t1, t2;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    cin >> t1 >> t2;
    t1--;
    t2--;
    edge[t2].push_back(t1);
    edge[t1].push_back(t2);
  }
  dfs_sz(0, -1);
  dfs_cnt(0, -1);
  for (int i = 0; i < n; i++) {
  }
  dfs(0, -1, 0, 0);
  cout << maxi << endl;
  return 0;
}

```