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>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb emplace_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define sz(x) (ll)x.size()
#define F first
#define S second
#define FOR(i,a,b) for(ll i=a;i<=b;++i)
#define ROF(i,a,b) for(ll i=a;i>=b;--i)
typedef long long ll;
using namespace std;

const int N = 200001;

int n;
int a[N], ans[N];
vector<int>g[N];
vector<int>dp;

void dfs(int u, int p){
    int pos = lower_bound(all(dp), a[u]) - dp.begin();
    int val = dp[pos];
    dp[pos] = a[u];
    ans[u] = lower_bound(all(dp), 1e9) - dp.begin() - 1;
    for(int v:g[u]){
        if(v != p)
            dfs(v, u);
    }
    dp[pos] = val;
}

int main(){
    IOS

    cin >> n;
    FOR(i,1,n){
        cin >> a[i];
    }
    FOR(i,1,n-1){
        int u, v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    dp = vector<int>(n+1,1e9);
    dp[0] = 0;
    dfs(1, 0);
    FOR(i,1,n){
        cout << ans[i] << endl;
    }

    return 0;
}