#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, a[N], depth[N];
long long Up[N], Down[N], Total_down[N], sum, sumDown[N];
vector<int> adj[N];
void input() {
  scanf("%d", &n);
  for (int i = (1), _b = (n); i <= _b; i++) {
    scanf("%d", &a[i]);
    sum += a[i];
  }
  for (int nhi = 0, _n = (n - 1); nhi < _n; nhi++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
void dfs1(int u, int par) {
  if (par != -1) depth[u] = depth[par] + 1;
  for (int v : adj[u]) {
    if (v == par) continue;
    dfs1(v, u);
    Down[u] += Down[v] + Total_down[v];
    Total_down[u] += Total_down[v];
    sumDown[u] += Down[v];
  }
  Total_down[u] += a[u];
}
void dfs2(int u, int par) {
  if (par != -1) {
    long long tmp = sum - Total_down[par] + a[par];
    Up[u] += Up[par] + tmp;
    Up[u] += (sumDown[par] - Down[u]) + 2 * (sum - Total_down[u] - tmp);
  }
  for (int v : adj[u]) {
    if (v == par) continue;
    dfs2(v, u);
  }
}
void solve() {
  dfs1(1, -1);
  dfs2(1, -1);
  long long res = 0;
  for (int i = (1), _b = (n); i <= _b; i++) res = max(res, Up[i] + Down[i]);
  printf("%I64d\n", res);
}
int main(int argc, char const *argv[]) {
  input();
  solve();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 int N = 2e5 + 5;
int n, a[N], depth[N];
long long Up[N], Down[N], Total_down[N], sum, sumDown[N];
vector<int> adj[N];
void input() {
  scanf("%d", &n);
  for (int i = (1), _b = (n); i <= _b; i++) {
    scanf("%d", &a[i]);
    sum += a[i];
  }
  for (int nhi = 0, _n = (n - 1); nhi < _n; nhi++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
void dfs1(int u, int par) {
  if (par != -1) depth[u] = depth[par] + 1;
  for (int v : adj[u]) {
    if (v == par) continue;
    dfs1(v, u);
    Down[u] += Down[v] + Total_down[v];
    Total_down[u] += Total_down[v];
    sumDown[u] += Down[v];
  }
  Total_down[u] += a[u];
}
void dfs2(int u, int par) {
  if (par != -1) {
    long long tmp = sum - Total_down[par] + a[par];
    Up[u] += Up[par] + tmp;
    Up[u] += (sumDown[par] - Down[u]) + 2 * (sum - Total_down[u] - tmp);
  }
  for (int v : adj[u]) {
    if (v == par) continue;
    dfs2(v, u);
  }
}
void solve() {
  dfs1(1, -1);
  dfs2(1, -1);
  long long res = 0;
  for (int i = (1), _b = (n); i <= _b; i++) res = max(res, Up[i] + Down[i]);
  printf("%I64d\n", res);
}
int main(int argc, char const *argv[]) {
  input();
  solve();
  return 0;
}

```