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;
using ll = long long;

const int maxn = 2e5+5, inf = 1e9+7;
int n, a[maxn], ans[maxn], lis[maxn];
vector<int> vs[maxn], seq;
vector<pair<int,int>> changes;


void dfs(int v, int p = -1){
	seq.emplace_back(a[v]);
	auto it = lower_bound(lis, lis+maxn, a[v]);
	changes.push_back({int(it-lis), *it});
	*it = min(*it, a[v]);
	ans[v] = lower_bound(lis, lis+maxn, inf)-1-lis;
	for(int e: vs[v]) if (e != p) dfs(e, v);
	seq.pop_back();
	pair<int,int> aux = changes.back();
	changes.pop_back();
	lis[aux.first] = aux.second;
}

int main(){
	scanf("%d", &n);
	fill(lis+1, lis+maxn, inf);
	lis[0] = 0;
	for(int i = 0; i<n; i++) scanf("%d", &a[i]);
	for(int i = 0; i<n-1; i++){
		int x, y;
		scanf("%d %d", &x, &y);
		vs[--x].emplace_back(--y);
		vs[y].emplace_back(x);
	}
	dfs(0);
	for(int i = 0; i<n; i++) printf("%d\n", ans[i]);

	return 0;
}