#include <bits/stdc++.h>
using namespace std;
int n, a[200010];
vector<int> G[200010];
long long res = 0;
long long tot;
int fa[200010];
long long f[200010], g[200010], sum[200010];
void DFS1(int u) {
  sum[u] = a[u];
  f[u] = 0;
  for (auto v : G[u])
    if (v != fa[u]) {
      fa[v] = u;
      DFS1(v);
      sum[u] += sum[v];
      f[u] += f[v] + sum[v];
    }
}
void DFS2(int u) {
  res = max(res, f[u] + g[u]);
  for (auto v : G[u])
    if (v != fa[u]) {
      g[v] = g[u] + f[u] - f[v] + tot - 2 * sum[v];
      DFS2(v);
    }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; ++i) G[i].clear();
    tot = 0;
    res = 0;
    g[1] = 0;
    for (int i = 1; i <= n; ++i) scanf("%d", a + i), tot += a[i];
    for (int i = 1, u, v; i < n; ++i) {
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    DFS1(1), DFS2(1);
    printf("%lld\n", res);
  }
  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;
int n, a[200010];
vector<int> G[200010];
long long res = 0;
long long tot;
int fa[200010];
long long f[200010], g[200010], sum[200010];
void DFS1(int u) {
  sum[u] = a[u];
  f[u] = 0;
  for (auto v : G[u])
    if (v != fa[u]) {
      fa[v] = u;
      DFS1(v);
      sum[u] += sum[v];
      f[u] += f[v] + sum[v];
    }
}
void DFS2(int u) {
  res = max(res, f[u] + g[u]);
  for (auto v : G[u])
    if (v != fa[u]) {
      g[v] = g[u] + f[u] - f[v] + tot - 2 * sum[v];
      DFS2(v);
    }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; ++i) G[i].clear();
    tot = 0;
    res = 0;
    g[1] = 0;
    for (int i = 1; i <= n; ++i) scanf("%d", a + i), tot += a[i];
    for (int i = 1, u, v; i < n; ++i) {
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    DFS1(1), DFS2(1);
    printf("%lld\n", res);
  }
  return 0;
}

```