#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, size = 1 << 20, mod = 998244353, inf = 2e9;
const long long INF = 1e15;
template <class o>
void qr(o &x) {
  char c = getchar();
  x = 0;
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
template <class o>
void qw(o x) {
  if (x / 10) qw(x / 10);
  putchar(x % 10 + '0');
}
template <class o>
void pr1(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar(' ');
}
template <class o>
void pr2(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar('\n');
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long power(long long a, long long b = mod - 2) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return c;
}
int n, a[N], dep[N];
long long s[N], f[N], ans;
vector<int> e[N];
void add(int x, int y) {
  e[x].push_back(y);
  e[y].push_back(x);
}
void dfs(int x, int fa = 0) {
  s[x] = a[x];
  f[x] = s[x] * dep[x];
  for (int y : e[x])
    if (y ^ fa) dep[y] = dep[x] + 1, dfs(y, x), f[x] += f[y], s[x] += s[y];
}
void DFS(int x, int fa = 0) {
  ans = max(ans, f[x]);
  for (int y : e[x])
    if (y ^ fa) f[y] = f[x] + s[1] - 2 * s[y], DFS(y, x);
}
void solve() {
  qr(n);
  for (int i = 1; i <= n; i++) qr(a[i]);
  for (int i = 1, x, y; i < n; i++) qr(x), qr(y), add(x, y);
  dfs(1);
  DFS(1);
  pr2(ans);
}
int main() {
  int T;
  solve();
}

### Prompt

Please create a solution in cpp 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 N = 2e5 + 10, size = 1 << 20, mod = 998244353, inf = 2e9;
const long long INF = 1e15;
template <class o>
void qr(o &x) {
  char c = getchar();
  x = 0;
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
template <class o>
void qw(o x) {
  if (x / 10) qw(x / 10);
  putchar(x % 10 + '0');
}
template <class o>
void pr1(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar(' ');
}
template <class o>
void pr2(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar('\n');
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long power(long long a, long long b = mod - 2) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return c;
}
int n, a[N], dep[N];
long long s[N], f[N], ans;
vector<int> e[N];
void add(int x, int y) {
  e[x].push_back(y);
  e[y].push_back(x);
}
void dfs(int x, int fa = 0) {
  s[x] = a[x];
  f[x] = s[x] * dep[x];
  for (int y : e[x])
    if (y ^ fa) dep[y] = dep[x] + 1, dfs(y, x), f[x] += f[y], s[x] += s[y];
}
void DFS(int x, int fa = 0) {
  ans = max(ans, f[x]);
  for (int y : e[x])
    if (y ^ fa) f[y] = f[x] + s[1] - 2 * s[y], DFS(y, x);
}
void solve() {
  qr(n);
  for (int i = 1; i <= n; i++) qr(a[i]);
  for (int i = 1, x, y; i < n; i++) qr(x), qr(y), add(x, y);
  dfs(1);
  DFS(1);
  pr2(ans);
}
int main() {
  int T;
  solve();
}

```