#include <bits/stdc++.h>
using namespace std;
const int maxn = 200100, maxm = 400100;
struct edge {
  int u, v;
  edge *next;
} epool[maxm], *edges[maxn];
int ecnt;
void add_edge(int u, int v) {
  edge *p = &epool[++ecnt], *q = &epool[++ecnt];
  p->u = u;
  p->v = v;
  p->next = edges[u];
  edges[u] = p;
  q->u = v;
  q->v = u;
  q->next = edges[v];
  edges[v] = q;
  return;
}
int n;
long long a[maxn], asum[maxn];
long long result;
long long dfs_1(int p, int q, long long depth) {
  long long distsum = a[p] * depth;
  asum[p] = a[p];
  for (edge *ep = edges[p]; ep; ep = ep->next)
    if (ep->v != q) {
      distsum += dfs_1(ep->v, p, depth + 1);
      asum[p] += asum[ep->v];
    }
  return distsum;
}
void dfs_2(int p, int q, long long distsum) {
  result = max(result, distsum);
  for (edge *ep = edges[p]; ep; ep = ep->next)
    if (ep->v != q) {
      distsum -= asum[ep->v];
      asum[p] -= asum[ep->v];
      distsum += asum[p];
      asum[ep->v] += asum[p];
      dfs_2(ep->v, p, distsum);
      asum[ep->v] -= asum[p];
      distsum -= asum[p];
      asum[p] += asum[ep->v];
      distsum += asum[ep->v];
    }
  return;
}
int main(int argc, char **argv) {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  for (int i = (1); i <= (n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
  }
  long long res = dfs_1(1, 0, 0);
  dfs_2(1, 0, res);
  printf("%lld\n", result);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 200100, maxm = 400100;
struct edge {
  int u, v;
  edge *next;
} epool[maxm], *edges[maxn];
int ecnt;
void add_edge(int u, int v) {
  edge *p = &epool[++ecnt], *q = &epool[++ecnt];
  p->u = u;
  p->v = v;
  p->next = edges[u];
  edges[u] = p;
  q->u = v;
  q->v = u;
  q->next = edges[v];
  edges[v] = q;
  return;
}
int n;
long long a[maxn], asum[maxn];
long long result;
long long dfs_1(int p, int q, long long depth) {
  long long distsum = a[p] * depth;
  asum[p] = a[p];
  for (edge *ep = edges[p]; ep; ep = ep->next)
    if (ep->v != q) {
      distsum += dfs_1(ep->v, p, depth + 1);
      asum[p] += asum[ep->v];
    }
  return distsum;
}
void dfs_2(int p, int q, long long distsum) {
  result = max(result, distsum);
  for (edge *ep = edges[p]; ep; ep = ep->next)
    if (ep->v != q) {
      distsum -= asum[ep->v];
      asum[p] -= asum[ep->v];
      distsum += asum[p];
      asum[ep->v] += asum[p];
      dfs_2(ep->v, p, distsum);
      asum[ep->v] -= asum[p];
      distsum -= asum[p];
      asum[p] += asum[ep->v];
      distsum += asum[ep->v];
    }
  return;
}
int main(int argc, char **argv) {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  for (int i = (1); i <= (n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
  }
  long long res = dfs_1(1, 0, 0);
  dfs_2(1, 0, res);
  printf("%lld\n", result);
  return 0;
}

```