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;
typedef long long ll;
int n,a[200005];
vector <int> v[200005];
set <int> s;
int ans[200005];

void dfs(int x,int p) {
	auto it = s.end();
	int val,val2,ch = 0;
	if(s.empty()) s.insert(a[x]);
	else {
		it--; val = *it;
		if(a[x] > val) ch = 1,s.insert(a[x]);
		else {
			it = s.lower_bound(a[x]);
			val2 = *it;
			s.erase(it);
			s.insert(a[x]);
			ch = 2;
		}
	}
	ans[x] = s.size();
	for(int i : v[x]) {
		if(i == p) continue;
		dfs(i,x);
	}
	if(!ch) s.erase(a[x]);
	if(ch == 1) s.erase(a[x]);
	if(ch == 2) {
		s.erase(a[x]);
		s.insert(val2);
	}
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for(int i = 1;i <= n;i++) cin >> a[i];
	for(int i = 1;i < n;i++) {
		int x,y; cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(1,-1);
	for(int i = 1;i <= n;i++) {
		cout << ans[i] << '\n';
	}
} 