#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
const int MOD = 998244353;
const double EPS = 1e-5;
struct star {
  int next, to;
};
int head[MAXN], n, len, en[MAXN];
star g[MAXN * 2];
long long int ans, a[MAXN], dp[MAXN], num[MAXN];
bool vis[MAXN];
queue<int> q;
void init() {
  memset(head, -1, sizeof(head));
  len = 0;
  memset(num, 0, sizeof(num));
  memset(en, 0, sizeof(en));
}
void add(int u, int v) {
  g[++len].to = v;
  g[len].next = head[u];
  head[u] = len;
}
long long int dfs(int now, int last) {
  if (now != 1 && en[now] == 1) {
    dp[now] = 0;
    num[now] = a[now];
    return 0;
  }
  long long int re = 0, sum = 0;
  for (int i = head[now]; i != -1; i = g[i].next) {
    if (g[i].to != last) {
      re += dfs(g[i].to, now);
      sum += num[g[i].to];
    }
  }
  re += sum;
  sum += a[now];
  dp[now] = re;
  num[now] = sum;
  return re;
}
void bfs(int st) {
  memset(vis, false, sizeof(vis));
  vis[st] = true;
  q.push(st);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = head[now]; i != -1; i = g[i].next) {
      if (!vis[g[i].to]) {
        vis[g[i].to] = true;
        q.push(g[i].to);
        long long int nowdp = dp[now];
        long long int nownum = num[now];
        nowdp = nowdp - dp[g[i].to] - num[g[i].to];
        nownum = nownum - num[g[i].to];
        dp[g[i].to] = dp[g[i].to] + nowdp + nownum;
        ans = max(ans, dp[g[i].to]);
        num[g[i].to] = num[g[i].to] + nownum;
      }
    }
  }
}
void ceshi1() {
  for (int i = 1; i <= n; ++i) {
    printf("%d:dp = %I64d  num = %I64d\n", i, dp[i], num[i]);
  }
}
void ceshi2() {
  printf("dpans:\n");
  for (int i = 1; i <= n; ++i) {
    printf("%d %I64d\n", i, dp[i]);
  }
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
    en[u]++;
    en[v]++;
  }
  dfs(1, 0);
  ans = dp[1];
  bfs(1);
  printf("%I64d\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;
const int MAXN = 2e5 + 10;
const int MOD = 998244353;
const double EPS = 1e-5;
struct star {
  int next, to;
};
int head[MAXN], n, len, en[MAXN];
star g[MAXN * 2];
long long int ans, a[MAXN], dp[MAXN], num[MAXN];
bool vis[MAXN];
queue<int> q;
void init() {
  memset(head, -1, sizeof(head));
  len = 0;
  memset(num, 0, sizeof(num));
  memset(en, 0, sizeof(en));
}
void add(int u, int v) {
  g[++len].to = v;
  g[len].next = head[u];
  head[u] = len;
}
long long int dfs(int now, int last) {
  if (now != 1 && en[now] == 1) {
    dp[now] = 0;
    num[now] = a[now];
    return 0;
  }
  long long int re = 0, sum = 0;
  for (int i = head[now]; i != -1; i = g[i].next) {
    if (g[i].to != last) {
      re += dfs(g[i].to, now);
      sum += num[g[i].to];
    }
  }
  re += sum;
  sum += a[now];
  dp[now] = re;
  num[now] = sum;
  return re;
}
void bfs(int st) {
  memset(vis, false, sizeof(vis));
  vis[st] = true;
  q.push(st);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = head[now]; i != -1; i = g[i].next) {
      if (!vis[g[i].to]) {
        vis[g[i].to] = true;
        q.push(g[i].to);
        long long int nowdp = dp[now];
        long long int nownum = num[now];
        nowdp = nowdp - dp[g[i].to] - num[g[i].to];
        nownum = nownum - num[g[i].to];
        dp[g[i].to] = dp[g[i].to] + nowdp + nownum;
        ans = max(ans, dp[g[i].to]);
        num[g[i].to] = num[g[i].to] + nownum;
      }
    }
  }
}
void ceshi1() {
  for (int i = 1; i <= n; ++i) {
    printf("%d:dp = %I64d  num = %I64d\n", i, dp[i], num[i]);
  }
}
void ceshi2() {
  printf("dpans:\n");
  for (int i = 1; i <= n; ++i) {
    printf("%d %I64d\n", i, dp[i]);
  }
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
    en[u]++;
    en[v]++;
  }
  dfs(1, 0);
  ans = dp[1];
  bfs(1);
  printf("%I64d\n", ans);
  return 0;
}

```