Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
int lis[200020],r=0,ans[200020],d[200020];
int ta[200020],nt[400040],to[400040];
int lb(int l,int r,int n,int *d){
  int m=(l+r)/2;
  if(l==r)return m;
  return n<=d[m]?lb(l,m,n,d):lb(m+1,r,n,d);
}
void f(int s,int t){//printf("%d %d\n",s,t);
  int rn,rp,i;
  ans[s]=r;
  for(i=ta[s];i+1;i=nt[i]){
    if(to[i]==t)continue;
    //int j;printf("lis %d:",s);
    //for(j=0;j<r;j++)printf("%d ",lis[j]);printf("\n");
    if(lis[r-1]<d[to[i]]){
      rn=-1;
      rp=-1;
      lis[r++]=d[to[i]];
    }
    else{
      rp=lb(0,r,d[to[i]],lis);
      rn=lis[rp];
      lis[rp]=d[to[i]];
    }
    f(to[i],s);
    if(rp+1)lis[rp]=rn;
    else    r--;
  }
}    
int main(){
  int n,i,a,b;
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&d[i+1]);
  for(i=0;i<n;i++)ta[i+1]=-1;
  for(i=0;i<n-1;i++){
    scanf("%d %d",&a,&b);
    nt[i    ]=ta[to[n-1+i]=a];
    nt[n-1+i]=ta[to[i    ]=b];
    ta[a]=i;
    ta[b]=n-1+i;
  }
  lis[r++]=d[1];
  f(1,0);
  for(i=0;i<n;i++)printf("%d\n",ans[i+1]);
  return 0;
}