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 rep(i,n) for(int i = 0; i < (int)(n); i++)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
using namespace std;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vs = vector<string>;
using vll = vector<long long>;
using vp = vector<pair<int, int>>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvll = vector<vector<long long>>;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

const int INF = 1001001001;
vvi to;
vi ans;
vi a;

void dfs(int v, vi& dp, int p=-1) {
    for (int u : to[v]) {
        if (u == p) continue;
        int idx = lower_bound(dp.begin(), dp.end(), a[u]) - dp.begin();
        int pre = dp[idx];
        dp[idx] = a[u];
        ans[u] = lower_bound(dp.begin(), dp.end(), INF) - dp.begin();
        dfs(u, dp, v);
        dp[idx] = pre;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n;
    cin >> n;
    a.resize(n);
    rep(i, n) cin >> a[i];
    to.resize(n);
    rep(i, n-1) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        to[u].pb(v);
        to[v].pb(u);
    }
    ans.resize(n);
    ans[0] = 1;

    vi dp(n, INF);
    dp[0] = a[0];
    dfs(0, dp);
    rep(i, n) cout << ans[i] << endl;
}