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;
const int maxN = 2e5 + 10;
int n, a[maxN], ans[maxN];
vector<int> edge[maxN], v;
void dfs(int x, int par) {
	int tmp = 0, pos, flag = 0;
	if(v.empty() || a[x] > v.back()) {
		v.push_back(a[x]);
	}else {
		flag = 1;
		pos = lower_bound(v.begin(), v.end(), a[x]) - v.begin();
		tmp = v[pos];
		v[pos] = a[x];
	}
	ans[x] = v.size(); 
	for(auto to : edge[x]) {
		if(to != par) {
			dfs(to, x);
		}
	}
	if(!flag) {
		v.pop_back();
	}else {
		v[pos] = tmp;
	}
}
int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%d", a + i);
	}
	for(int i = 0; i < n - 1; i++) {
		int u, v;
		scanf("%d%d", &u, &v); 
		--u, --v;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	dfs(0, -1);
	for(int i = 0; i < n; i++) {
		printf("%d\n", ans[i]);
	}
	return 0;
}