#include <bits/stdc++.h>
using namespace std;
struct node {
  int next, to;
  long long dis;
} edge[2000010];
int head[2000010], num_edge;
void add_edge(int from, int to, long long dis) {
  edge[++num_edge].next = head[from];
  edge[num_edge].to = to;
  edge[num_edge].dis = dis;
  head[from] = num_edge;
}
long long siz[2000010], dis[2000010], f[2000010], ans = -9223372036854775807,
                                                  sum;
int n, a, b;
void dfs(int fr, int fa) {
  for (int i = head[fr]; i; i = edge[i].next) {
    int ca = edge[i].to;
    if (ca == fa) continue;
    dfs(ca, fr);
    siz[fr] += siz[ca];
    dis[fr] += dis[ca] + edge[i].dis * siz[ca];
  }
}
void dd(int fr, int fa) {
  ans = max(ans, f[fr]);
  for (int i = head[fr]; i; i = edge[i].next) {
    int ca = edge[i].to;
    if (ca == fa) continue;
    f[ca] = f[fr] + (sum - siz[ca]) * edge[i].dis - siz[ca] * edge[i].dis;
    dd(ca, fr);
  }
}
int main(int argc, const char *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> siz[i], sum += siz[i];
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    add_edge(a, b, 1);
    add_edge(b, a, 1);
  }
  dfs(1, 0);
  f[1] = dis[1];
  dd(1, 0);
  if (ans == -9223372036854775807)
    puts("0");
  else
    cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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;
struct node {
  int next, to;
  long long dis;
} edge[2000010];
int head[2000010], num_edge;
void add_edge(int from, int to, long long dis) {
  edge[++num_edge].next = head[from];
  edge[num_edge].to = to;
  edge[num_edge].dis = dis;
  head[from] = num_edge;
}
long long siz[2000010], dis[2000010], f[2000010], ans = -9223372036854775807,
                                                  sum;
int n, a, b;
void dfs(int fr, int fa) {
  for (int i = head[fr]; i; i = edge[i].next) {
    int ca = edge[i].to;
    if (ca == fa) continue;
    dfs(ca, fr);
    siz[fr] += siz[ca];
    dis[fr] += dis[ca] + edge[i].dis * siz[ca];
  }
}
void dd(int fr, int fa) {
  ans = max(ans, f[fr]);
  for (int i = head[fr]; i; i = edge[i].next) {
    int ca = edge[i].to;
    if (ca == fa) continue;
    f[ca] = f[fr] + (sum - siz[ca]) * edge[i].dis - siz[ca] * edge[i].dis;
    dd(ca, fr);
  }
}
int main(int argc, const char *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> siz[i], sum += siz[i];
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    add_edge(a, b, 1);
    add_edge(b, a, 1);
  }
  dfs(1, 0);
  f[1] = dis[1];
  dd(1, 0);
  if (ans == -9223372036854775807)
    puts("0");
  else
    cout << ans << endl;
}

```