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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int  N=2e5+10;

int n,w[N],v[N],len,ans[N];
struct edge{int y,next;}a[N<<1]; int Len,last[N];
void ins(int x,int y) {a[++Len]=(edge){y,last[x]};last[x]=Len;}

void dfs(int x,int f) {
	int t=len+1,val=0; bool flag=0;
	if(w[len]<v[x]) w[++len]=v[x],flag=1;
	else t=lower_bound(w+1,w+len+1,v[x])-w,val=w[t],w[t]=v[x];
	ans[x]=len;
	for(int k=last[x],y;k;k=a[k].next) 
		if((y=a[k].y)^f) dfs(y,x);
	w[t]=val;
	len-=flag;
}

int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&v[i]);
	for(int i=1,x,y;i<n;i++)
		scanf("%d%d",&x,&y),ins(x,y),ins(y,x);
	dfs(1,0);for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
	return 0;
}
