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 <bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=2e5+10;
int n;
vector<int> g[maxn],lsh;
int a[maxn],ans[maxn],rec[maxn],pre[maxn];
vector<int> f[maxn],h;
void solve(int x,int fa)
{
	int pos=lower_bound(h.begin(),h.end(),a[x])-h.begin();
	f[pos].push_back(a[x]);
	h[pos]=a[x];
	rec[x]=pos;
	ans[x]=lower_bound(h.begin(),h.end(),inf)-h.begin();
	
	for(int i=0;i<g[x].size();i++)
	{
		int to=g[x][i];
		if(to==fa)continue;
		solve(to,x);
	}
	
	f[pos].pop_back();
	h[pos]=f[pos].back();
}
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		lsh.push_back(a[i]);
	}
	sort(lsh.begin(),lsh.end());
	lsh.erase(unique(lsh.begin(),lsh.end()),lsh.end());
	for(int i=0;i<n;i++)
	{
		a[i]=lower_bound(lsh.begin(),lsh.end(),a[i])-lsh.begin();
	}
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		u--,v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	f[0].push_back(-1);
	h.push_back(-1);
	for(int i=1;i<=n+1;i++)
	{
		f[i].push_back(inf);
		h.push_back(inf);
	}
	solve(0,-1);
	for(int i=0;i<n;i++)
	{
		printf("%d\n",ans[i]-1);
	}
	return 0;
}