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.

Example

Input

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


Output

1
2
3
3
4
4
5
2
2
3
#include<bits/stdc++.h>
using namespace std;

#define ran 202202

int n;
int a[ran];
vector<int> e[ran];

int f[ran];
int len;

int ans[ran];

void dfs(int cur, int par) {
	int val = a[cur];
	int it = lower_bound(f, f+len, val) - f;
	int tmp = f[it];
	f[it] = val;
	
	bool inc = false;
	if(it>=len) { len++; inc = true; }
	
	ans[cur] = len;
	
	for(int i=0;i<(int)e[cur].size();i++) {
		int nxt = e[cur][i];
		if(nxt==par) continue;
		dfs(nxt, cur);
	}
	
	f[it] = tmp;
	if (inc) len--;
}

int main() {
	scanf("%d", &n);
	for(int i=1;i<=n;i++) scanf("%d", &a[i]);
	for(int i=1;i<n;i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	
	len = 0;
	dfs(1, -1);
	
	for(int i=1;i<=n;i++) printf("%d\n", ans[i]);
	
	return 0;
}