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 N 200005

int len, n, ans[N], c[N], a[N], u, v, i;
vector<int> b[N];

void dfs(int u,int v)
{
	int i, x, d;
	pair<int, pair<int,int> > flag;
	
	for (i=0; i<b[u].size(); i++) {
		x=b[u][i];
		if (x!=v) {
			if (a[x]>c[len]) {
				c[++len]=a[x];
				flag=make_pair(1,make_pair(0,0));
			}
			else {
			    d=lower_bound(c+1,c+len+1,a[x])-c;
			    flag=make_pair(2,make_pair(d,c[d]));
			    c[d]=a[x];
			}
			ans[x]=len;
			dfs(x,u);
			if (flag.first==1) len--;
			else {
				c[flag.second.first]=flag.second.second;
			}
		}
	}
}

int main (void)
{
	scanf("%d",&n);
	for (i=1; i<=n; i++) scanf("%d",&a[i]);
	for (i=1; i<n; i++) {
		scanf("%d%d",&u,&v);
		b[u].push_back(v),b[v].push_back(u);
	}
	c[1]=a[1]; len=1; ans[1]=len; dfs(1,0);
	for (i=1; i<=n; i++) printf("%d\n",ans[i]);
	return 0;
}