#include <bits/stdc++.h>
using namespace std;
const int N = 5e5;
struct E {
  int t, next;
} e[N];
int head[N];
long long sum[N], v[N];
int n, cnt;
long long all, ans, tot;
void add(int u, int v) {
  e[cnt].t = v;
  e[cnt].next = head[u];
  head[u] = cnt++;
}
long long dfs1(int now, int f) {
  long long s = 0;
  for (int i = head[now]; ~i; i = e[i].next) {
    if (e[i].t == f) continue;
    s += dfs1(e[i].t, now);
  }
  all += s;
  return sum[now] = v[now] + s;
}
void dfs2(int now, long long nv, int f) {
  if (ans < nv) ans = nv;
  for (int i = head[now]; ~i; i = e[i].next) {
    if (e[i].t == f) continue;
    long long next = nv - 2 * sum[e[i].t] + tot;
    dfs2(e[i].t, next, now);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  memset(sum, 0, sizeof(sum));
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &v[i]);
    tot += v[i];
  }
  int u, v;
  cnt = ans = 0;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs1(1, -1);
  dfs2(1, all, -1);
  printf("%lld\n", ans);
}

### Prompt

Please provide a CPP coded solution 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 = 5e5;
struct E {
  int t, next;
} e[N];
int head[N];
long long sum[N], v[N];
int n, cnt;
long long all, ans, tot;
void add(int u, int v) {
  e[cnt].t = v;
  e[cnt].next = head[u];
  head[u] = cnt++;
}
long long dfs1(int now, int f) {
  long long s = 0;
  for (int i = head[now]; ~i; i = e[i].next) {
    if (e[i].t == f) continue;
    s += dfs1(e[i].t, now);
  }
  all += s;
  return sum[now] = v[now] + s;
}
void dfs2(int now, long long nv, int f) {
  if (ans < nv) ans = nv;
  for (int i = head[now]; ~i; i = e[i].next) {
    if (e[i].t == f) continue;
    long long next = nv - 2 * sum[e[i].t] + tot;
    dfs2(e[i].t, next, now);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  memset(sum, 0, sizeof(sum));
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &v[i]);
    tot += v[i];
  }
  int u, v;
  cnt = ans = 0;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs1(1, -1);
  dfs2(1, all, -1);
  printf("%lld\n", ans);
}

```