Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices, whose i-th edge connects Vertex u_i and Vertex v_i.
Vertex i has an integer a_i written on it.
For every integer k from 1 through N, solve the following problem:

We will make a sequence by lining up the integers written on the vertices along the shortest path from Vertex 1 to Vertex k, in the order they appear. Find the length of the longest increasing subsequence of this sequence.

Here, the longest increasing subsequence of a sequence A of length L is the subsequence A_{i_1} , A_{i_2} , ... , A_{i_M} with the greatest possible value of M such that 1 \leq i_1 < i_2 < ... < i_M \leq L and A_{i_1} < A_{i_2} < ... < A_{i_M}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq a_i \leq 10^9

1 \leq u_i , v_i \leq N

u_i \neq v_i

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
u_1 v_1
u_2 v_2
:
u_{N-1} v_{N-1}

Output

Print N lines. The k-th line, print the length of the longest increasing subsequence of the sequence obtained from the shortest path from Vertex 1 to Vertex k.

Sample Input 1

10
1 2 5 3 4 6 7 3 2 4
1 2
2 3
3 4
4 5
3 6
6 7
1 8
8 9
9 10

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define RI register int
typedef long long LL;

#define FILEIO(name) freopen(name".in", "r", stdin), freopen(name".out", "w", stdout);

using namespace std;

char buf[1000000], *p1 = buf, *p2 = buf;
inline char gc() {
  if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
  return p1 == p2 ? EOF : *(p1++);
}
template <class T> inline void read(T &n) {
  n = 0; RI ch = gc(), f;
  while ((ch < '0' || ch > '9') && ch != '-') ch = gc();
  f = (ch == '-' ? ch = gc(), -1 : 1);
  while (ch >= '0' && ch <= '9') n = n * 10 + (ch ^ 48), ch = gc();
  n *= f;
}

int const MAXN = 2e5 + 5;
int a[MAXN], Ans[MAXN];
int top;
int stk[MAXN];
struct Edges { int to, next; } e[MAXN * 2];
int head[MAXN], tot;

inline void addedge(int from, int to) {
  e[++tot] = (Edges){to, head[from]};
  head[from] = tot;
  e[++tot] = (Edges){from, head[to]};
  head[to] = tot;
}

void Dfs(int now, int fa) {
  int pos = lower_bound(stk + 1, stk + 1 + top, a[now]) - stk;
  int curstk = stk[pos], curtop = top;
  stk[pos] = a[now];
  top = max(top, pos);
  Ans[now] = top;
  for (RI i = head[now]; i; i = e[i].next)
    if (e[i].to != fa)
      Dfs(e[i].to, now);
  top = curtop;
  stk[pos] = curstk;
}

int main() {
  
#ifdef LOCAL
  FILEIO("a");
#endif

  int n; read(n);
  for (RI i = 1; i <= n; ++i) read(a[i]);
  for (RI i = 1, x, y; i < n; ++i)
    read(x), read(y), addedge(x, y);
  memset(stk, 0x7f, sizeof(stk));
  Dfs(1, 0);
  for (RI i = 1; i <= n; ++i)
    printf("%d\n", Ans[i]);

  return 0;
}

// created by Daniel yuan