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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=200007;
struct node{int mx,ls,rs;}t[N*350];
vector<int>e[N];int n,m,tot,ans[N],a[N],h[N],root[N];
int read(){int x=0,c=getchar();while(isspace(c))c=getchar();while(isdigit(c))(x*=10)+=c&15,c=getchar();return x;}
#define mid ((l+r)/2)
void build(int&p,int l,int r)
{
    p=++tot;
    if(l^r) build(t[p].ls,l,mid),build(t[p].rs,mid+1,r);
}
int query(int p,int l,int r,int L,int R)
{
    if(L>R||l>R||L>r) return 0;
    if(L<=l&&r<=R) return t[p].mx;
    return max(query(t[p].ls,l,mid,L,R),query(t[p].rs,mid+1,r,L,R));
}
void update(int&p,int pre,int l,int r,int x,int v)
{
    p=++tot,t[p].mx=max(t[pre].mx,v);
    if(l==r) return ;
    if(x<=mid) update(t[p].ls,t[pre].ls,l,mid,x,v),t[p].rs=t[pre].rs;
    else update(t[p].rs,t[pre].rs,mid+1,r,x,v),t[p].ls=t[pre].ls;
}
void dfs(int u,int fa)
{
    ans[u]=query(root[fa],1,m,1,a[u]-1)+1;
    update(root[u],root[fa],1,m,a[u],ans[u]);
    for(int v:e[u]) if(v^fa) dfs(v,u);
}
#undef mid
int main()
{
    n=read();
    for(int i=1;i<=n;++i) h[i]=a[i]=read();
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),e[u].push_back(v),e[v].push_back(u);
    sort(h+1,h+n+1),m=unique(h+1,h+n+1)-h-1;
    for(int i=1;i<=n;++i) a[i]=std::lower_bound(h+1,h+m+1,a[i])-h;
    build(root[0],1,m),dfs(1,0);
    for(int i=1;i<=n;++i) printf("%d\n",ans[i]);
}