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;
using ll = long long;
using PII = pair<ll, ll>;
#define FOR(i, a, n) for(ll i=(ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
template<typename T> void chmin(T &a, const T &b) { a = min(a, b); }
template<typename T> void chmax(T &a, const T &b) { a = max(a, b); }
struct FastIO {FastIO() { cin.tie(0); ios::sync_with_stdio(0); }}fastiofastio;
#ifdef DEBUG
#include "../program_contest_library/memo/dump.hpp"
#else
#define dump(...)
#endif
constexpr ll INF = 1LL<<60;

int main() {
    ll n;
    cin >> n;
    vector<ll> a(n);
    REP(i, n) cin >> a[i];
    vector<vector<ll>> g(n);
    REP(i, n-1) {
        ll u, v;
        cin >> u >> v;
        u--, v--;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    vector<ll> vs(a);
    sort(ALL(vs));
    vs.erase(unique(ALL(vs)), vs.end());
    REP(i, n) a[i] = lower_bound(ALL(vs), a[i]) - vs.begin();

    vector<ll> ans(n);
    auto dfs = [&](auto &&self, ll v, ll p, vector<ll> &dp) -> void {
        ll idx = lower_bound(ALL(dp), a[v]) - dp.begin();
        ll hold = dp[idx];
        dp[idx] = a[v];
        ans[v] = lower_bound(ALL(dp), INF) - dp.begin();
        for(auto to: g[v]) if(to!=p) {
            self(self, to, v, dp);
        }
        dp[idx] = hold;
    };
    vector<ll> dp(n, INF);
    dfs(dfs, 0, -1, dp);

    REP(i, n) cout << ans[i] << "\n";

    return 0;
}