#include <bits/stdc++.h>
using namespace std;
long long Head[400000 + 5], Cnt, N, A[400000 + 5], sigma, F[400000 + 5],
    S[400000 + 5];
struct Edge {
  long long Next, To;
} E[400000 << 2];
void Add_Edge(long long X, long long Y) {
  E[++Cnt].Next = Head[X];
  E[Cnt].To = Y, Head[X] = Cnt;
}
void DFS1(long long X, long long Fa) {
  S[X] = A[X];
  for (long long I = Head[X]; I; I = E[I].Next) {
    long long Y = E[I].To;
    if (Y == Fa) continue;
    DFS1(Y, X), S[X] += S[Y];
    F[X] += F[Y] + S[Y];
  }
}
void DFS2(long long X, long long Fa) {
  for (long long I = Head[X]; I; I = E[I].Next) {
    long long Y = E[I].To;
    if (Y == Fa) continue;
    F[Y] = F[X] - S[Y] + (sigma - S[Y]);
    DFS2(Y, X);
  }
}
signed main() {
  scanf("%lld", &N);
  for (long long I = 1; I <= N; I++) scanf("%lld", &A[I]), sigma += A[I];
  for (long long I = 1; I < N; I++) {
    long long U, V;
    scanf("%lld%lld", &U, &V);
    Add_Edge(U, V), Add_Edge(V, U);
  }
  DFS1(1, 0), DFS2(1, 0);
  long long pos = 0, v = 0;
  for (long long I = 1; I <= N; I++)
    if (v < F[I]) v = F[I], pos = I;
  printf("%lld\n", F[pos]);
  return 0;
}

### Prompt

Generate 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;
long long Head[400000 + 5], Cnt, N, A[400000 + 5], sigma, F[400000 + 5],
    S[400000 + 5];
struct Edge {
  long long Next, To;
} E[400000 << 2];
void Add_Edge(long long X, long long Y) {
  E[++Cnt].Next = Head[X];
  E[Cnt].To = Y, Head[X] = Cnt;
}
void DFS1(long long X, long long Fa) {
  S[X] = A[X];
  for (long long I = Head[X]; I; I = E[I].Next) {
    long long Y = E[I].To;
    if (Y == Fa) continue;
    DFS1(Y, X), S[X] += S[Y];
    F[X] += F[Y] + S[Y];
  }
}
void DFS2(long long X, long long Fa) {
  for (long long I = Head[X]; I; I = E[I].Next) {
    long long Y = E[I].To;
    if (Y == Fa) continue;
    F[Y] = F[X] - S[Y] + (sigma - S[Y]);
    DFS2(Y, X);
  }
}
signed main() {
  scanf("%lld", &N);
  for (long long I = 1; I <= N; I++) scanf("%lld", &A[I]), sigma += A[I];
  for (long long I = 1; I < N; I++) {
    long long U, V;
    scanf("%lld%lld", &U, &V);
    Add_Edge(U, V), Add_Edge(V, U);
  }
  DFS1(1, 0), DFS2(1, 0);
  long long pos = 0, v = 0;
  for (long long I = 1; I <= N; I++)
    if (v < F[I]) v = F[I], pos = I;
  printf("%lld\n", F[pos]);
  return 0;
}

```