#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
inline long long readll() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
template <typename T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
const double PI = acos((double)-1.0);
const double EPS = 1e-6;
inline int fcmp(double a, double b) {
  return (fabs(a - b) < EPS) ? 0 : (a < b ? -1 : 1);
}
const int INFINT = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const long long MOD1 = (long long)(1e9 + 7), MOD2 = 998244353;
const int maxn = 200100;
int n;
long long subw[maxn];
long long szw[maxn];
long long ans[maxn];
long long val[maxn];
vector<int> edge[maxn];
void init() {
  for (int i = 0; i <= n; ++i) {
    edge[i].clear();
    val[i] = 0;
  }
}
void dfs1(int u, int pre) {
  szw[u] = val[u];
  subw[u] = 0ll;
  for (auto &v : edge[u]) {
    if (v == pre) continue;
    dfs1(v, u);
    szw[u] += szw[v];
    subw[u] += subw[v] + szw[v];
  }
}
void dfs2(int u, int pre) {
  for (auto &v : edge[u]) {
    if (v == pre) continue;
    ans[v] = ans[u] + (szw[1] - szw[v]) - szw[v];
    dfs2(v, u);
  }
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; ++i) scanf("%lld", val + i);
  for (int i = 2, u, v; i <= n; ++i) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs1(1, 0);
  ans[1] = subw[1];
  dfs2(1, 0);
  long long mx = 0;
  for (int i = 1; i <= n; ++i) mx = max(mx, ans[i]);
  printf("%lld\n", mx);
}

### 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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
inline long long readll() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
template <typename T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
const double PI = acos((double)-1.0);
const double EPS = 1e-6;
inline int fcmp(double a, double b) {
  return (fabs(a - b) < EPS) ? 0 : (a < b ? -1 : 1);
}
const int INFINT = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const long long MOD1 = (long long)(1e9 + 7), MOD2 = 998244353;
const int maxn = 200100;
int n;
long long subw[maxn];
long long szw[maxn];
long long ans[maxn];
long long val[maxn];
vector<int> edge[maxn];
void init() {
  for (int i = 0; i <= n; ++i) {
    edge[i].clear();
    val[i] = 0;
  }
}
void dfs1(int u, int pre) {
  szw[u] = val[u];
  subw[u] = 0ll;
  for (auto &v : edge[u]) {
    if (v == pre) continue;
    dfs1(v, u);
    szw[u] += szw[v];
    subw[u] += subw[v] + szw[v];
  }
}
void dfs2(int u, int pre) {
  for (auto &v : edge[u]) {
    if (v == pre) continue;
    ans[v] = ans[u] + (szw[1] - szw[v]) - szw[v];
    dfs2(v, u);
  }
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; ++i) scanf("%lld", val + i);
  for (int i = 2, u, v; i <= n; ++i) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs1(1, 0);
  ans[1] = subw[1];
  dfs2(1, 0);
  long long mx = 0;
  for (int i = 1; i <= n; ++i) mx = max(mx, ans[i]);
  printf("%lld\n", mx);
}

```