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;
const int N = 2e5+10;

int n,c[N],f[N],top,head[N],ans[N];
struct zjz{
	int next,to;
}a[N << 1];

inline void add(int fr,int t){
	a[++top].to = t;
	a[top].next = head[fr];
	head[fr] = top;
	a[++top].to = fr;
	a[top].next = head[t];
	head[t] = top;
}

inline void dfs(int x,int fa){
	int *t = lower_bound(f,f+n+1,c[x]) , v = *t;
	*t = c[x] , ans[x] = lower_bound(f,f+n+1,0x7f7f7f7f)-f;
	for(int i=head[x];i != -1;i = a[i].next)
		if(a[i].to != fa)
			dfs(a[i].to,x);
	*t = v;
}

int main()
{
	memset(head,-1,sizeof(head));
	memset(f,0x7f,sizeof(f));
	int fr,t;
	scanf("%d",&n);
	for(int i=1;i <= n;i++) scanf("%d",&c[i]);
	for(int i=1;i < n;i++) scanf("%d%d",&fr,&t) , add(fr,t);
	dfs(1,0);
	for(int i=1;i <= n;i++) printf("%d\n",ans[i]);
	return 0;
} 