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 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct node{
	int pre, to;
}edge[400010];
int head[200010], tot;
int n, a[200010];
int low[200010], cnt;
int ans[200010];
void add(int u, int v) {
	edge[++tot] = node{head[u], v};
	head[u] = tot;
}
void dfs(int x, int fa) {
	ans[x] = cnt;
	for (int i = head[x]; i; i = edge[i].pre) {
		int y = edge[i].to;
		if (y != fa) {
			if (low[cnt] < a[y]) {
				int b = cnt;
				low[++cnt] = a[y];
				dfs(y, x);
				cnt = b;
			} else {
				int p = lower_bound(low + 1, low + cnt + 1, a[y]) - low;
				int v = low[p];
				low[p] = a[y];
				dfs(y, x);
				low[p] = v;
			}
		}
	}
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		add(u, v);
		add(v, u);
	}
	low[++cnt] = a[1];
	ans[1] = cnt;
	dfs(1, 0);
	for (int i = 1; i <= n; i++) {
		cout << ans[i] << "\n";
	}
	return 0;
}