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;
///////////////////////////////////////////
const long long int INF = 1LL<<60;
const long long int Mod = 1000000007;
using ll = long long int; using ci = const int;
using vi = vector<int>;  using Vi = vector<long long int>;
using P = pair<int, int>;  using PLL = pair<ll, ll>;
using matrix = vector<vector<ll>>;
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define rep(i,N) for(ll i = 0; i < (ll)N; i++)
#define repi(i,a,b) for(ll i = ll(a); i < ll(b); ++i)
template<class T>bool chmax(T &former, const T &b) { if (former<b) { former=b; return true; } return false; }
template<class T>bool chmin(T &former, const T &b) { if (b<former) { former=b; return true; } return false; }
template<class T>T sqar(T x){ return x*x; }//sqrt(x)は平方根;
#define Sort(v) std::sort(v.begin(), v.end(), std::greater<decltype(v[0])>()) //降順でVをソート
#define p_queue(v) priority_queue<v, vector<v>, greater<v> >
template<class T> inline void princ(T x){cout<<x<<" ";}; 
template<class T> inline void print(T x){cout<<x<<"\n";};
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
///////////////////////////////////////////////////////////////////////////////////

struct edge{/*重み付き,有向*/
    int to , cost;/*toは接続先*/
    edge(ll to_ ,ll cost_ = 1){ to=to_; cost=cost_; };/*全てのcost=1 <=> 無向グラフ*/
};
bool operator<(edge e1 ,edge e2){return e1.cost < e2.cost; }
bool operator>(edge e1 ,edge e2){return e1.cost > e2.cost; }
using graph =  vector< vector<edge> >;

ll n;
Vi a;
graph g;
Vi as;

void dfs(ll idn,ll prev,Vi& dp){
    ll tmp = lower_bound(all(dp),a[idn]) - dp.begin();
    ll mem = dp[tmp];
    dp[tmp] = a[idn];
    as[idn] = lower_bound(all(dp),INF)-dp.begin();
    rep(i,g[idn].size()){
        edge e = g[idn][i];
        if(e.to != prev){
            dfs(e.to,idn,dp);
        }
    }
  dp[tmp] = mem;
}

int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    std::cout<<std::fixed<<std::setprecision(30);
    cin>>n;
    g.resize(n);
    a.resize(n);
    as.resize(n);
    rep(i,n){
        cin>>a[i];
    }
    rep(i,n-1){
        ll in,inp;
        cin>>in>>inp;
        in--;
        inp--;
        g[in].pb(edge(inp));
        g[inp].pb(edge(in));
    }
    Vi tp;
    tp.resize(n,INF);
    dfs(0,-1,tp);
    rep(i,as.size()){
        print(as[i]);
    }
    return 0;
}