#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const long long INF = 3e9 + 7;
const int maxn = 2e6 + 6;
const int mod = 10007;
const double eps = 1e-3;
inline bool read(long long &num) {
  char in;
  bool IsN = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else
    num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') {
    num *= 10, num += in - '0';
  }
  if (IsN) num = -num;
  return true;
}
long long n, m;
long long num[maxn];
vector<int> v[maxn];
long long rt[maxn];
long long dp[maxn];
long long sz[maxn];
long long sum = 0;
void dfs(int u, int fa) {
  sz[u] = num[u];
  for (int e : v[u]) {
    if (e == fa) continue;
    dfs(e, u);
    sz[u] += sz[e];
    dp[u] += dp[e] + sz[e];
  }
}
void dfs2(int u, int fa) {
  if (u != fa) {
    rt[u] = dp[u] + (rt[fa] - sz[u] - dp[u]) + sum - sz[u];
  }
  for (int e : v[u]) {
    if (e == fa) continue;
    dfs2(e, u);
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(num[i]);
    sum += num[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 1);
  rt[1] = dp[1];
  dfs2(1, 1);
  long long maxl = 0;
  for (int i = 1; i <= n; i++) maxl = max(maxl, rt[i]);
  printf("%lld\n", maxl);
  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>
#pragma GCC optimize(2)
using namespace std;
const long long INF = 3e9 + 7;
const int maxn = 2e6 + 6;
const int mod = 10007;
const double eps = 1e-3;
inline bool read(long long &num) {
  char in;
  bool IsN = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else
    num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') {
    num *= 10, num += in - '0';
  }
  if (IsN) num = -num;
  return true;
}
long long n, m;
long long num[maxn];
vector<int> v[maxn];
long long rt[maxn];
long long dp[maxn];
long long sz[maxn];
long long sum = 0;
void dfs(int u, int fa) {
  sz[u] = num[u];
  for (int e : v[u]) {
    if (e == fa) continue;
    dfs(e, u);
    sz[u] += sz[e];
    dp[u] += dp[e] + sz[e];
  }
}
void dfs2(int u, int fa) {
  if (u != fa) {
    rt[u] = dp[u] + (rt[fa] - sz[u] - dp[u]) + sum - sz[u];
  }
  for (int e : v[u]) {
    if (e == fa) continue;
    dfs2(e, u);
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(num[i]);
    sum += num[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 1);
  rt[1] = dp[1];
  dfs2(1, 1);
  long long maxl = 0;
  for (int i = 1; i <= n; i++) maxl = max(maxl, rt[i]);
  printf("%lld\n", maxl);
  return 0;
}

```