#include <bits/stdc++.h>
using namespace std;
long long n, x, y;
long long a[200010], d[200010], ch[200010];
vector<long long> v[200010], c;
long long ic[200010];
long long b[200010] = {};
long long tmp = 0, ans = 0;
long long s = 0;
void upd(long long x, long long v) {
  while (x <= 200005) {
    b[x] += v;
    x += (x & -x);
  }
}
long long get(long long x) {
  if (!x) return 0;
  long long r = 0;
  while (x) {
    r += b[x];
    x -= (x & -x);
  }
  return r;
}
long long dfs(long long x, long long p, long long dp) {
  tmp += dp * a[x];
  d[x] = dp;
  c.push_back(x);
  ic[x] = c.size();
  ch[x] = 1;
  for (auto e : v[x]) {
    if (e == p) continue;
    ch[x] += dfs(e, x, dp + 1);
  }
  return ch[x];
}
void dfs2(long long x, long long p, long long t) {
  ans = max(ans, t);
  for (auto e : v[x]) {
    if (e == p) continue;
    tmp = get(ic[e] + ch[e] - 1) - get(ic[e] - 1);
    dfs2(e, x, t - tmp + s - tmp);
  }
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    s += a[i];
  }
  for (long long i = 1; i < n; ++i) {
    scanf("%lld %lld", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0, 0);
  for (long long i = 0; i < n; ++i) {
    upd(i + 1, a[c[i]]);
  }
  dfs2(1, 0, tmp);
  printf("%lld\n", ans);
  return 0;
}

### 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;
long long n, x, y;
long long a[200010], d[200010], ch[200010];
vector<long long> v[200010], c;
long long ic[200010];
long long b[200010] = {};
long long tmp = 0, ans = 0;
long long s = 0;
void upd(long long x, long long v) {
  while (x <= 200005) {
    b[x] += v;
    x += (x & -x);
  }
}
long long get(long long x) {
  if (!x) return 0;
  long long r = 0;
  while (x) {
    r += b[x];
    x -= (x & -x);
  }
  return r;
}
long long dfs(long long x, long long p, long long dp) {
  tmp += dp * a[x];
  d[x] = dp;
  c.push_back(x);
  ic[x] = c.size();
  ch[x] = 1;
  for (auto e : v[x]) {
    if (e == p) continue;
    ch[x] += dfs(e, x, dp + 1);
  }
  return ch[x];
}
void dfs2(long long x, long long p, long long t) {
  ans = max(ans, t);
  for (auto e : v[x]) {
    if (e == p) continue;
    tmp = get(ic[e] + ch[e] - 1) - get(ic[e] - 1);
    dfs2(e, x, t - tmp + s - tmp);
  }
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    s += a[i];
  }
  for (long long i = 1; i < n; ++i) {
    scanf("%lld %lld", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0, 0);
  for (long long i = 0; i < n; ++i) {
    upd(i + 1, a[c[i]]);
  }
  dfs2(1, 0, tmp);
  printf("%lld\n", ans);
  return 0;
}

```