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 <stdio.h>

typedef struct List {
	struct List *next;
	int v;
} list;

int N, a[200001], ans[200001] = {}, dp[200001], flag[200001] = {};
list *adj[200001] = {}, e[400001];

void DFS(int u, int k)
{
	if (flag[u] == 1) return;
	else flag[u] = 1;
	
	int l = 0, r = k + 1, prev;
	while (l < r) {
		if (dp[(l+r)/2] < a[u]) l = (l + r) / 2 + 1;
		else r = (l + r) / 2;
	}
	prev = dp[l];
	dp[l] = a[u];
	if (l == k + 1) ans[u] = k + 1;
	else ans[u] = k;
	
	list *p;
	for (p = adj[u]; p != NULL; p = p->next) DFS(p->v, ans[u]);
	
	dp[l] = prev;
}

int main()
{
	int i, u, w;
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(a[i]));
	for (i = 0; i < N - 1; i++) {
		scanf("%d %d", &u, &w);
		e[i*2].v = w;
		e[i*2+1].v = u;
		e[i*2].next = adj[u];
		e[i*2+1].next = adj[w];
		adj[u] = &(e[i*2]);
		adj[w] = &(e[i*2+1]);
	}
	
	ans[1] = 1;
	dp[0] = 0;
	dp[1] = a[1];
	DFS(1, ans[1]);
	for (i = 1; i <= N; i++) printf("%d\n", ans[i]);
	fflush(stdout);
	return 0;
}