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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define ALL(a) (a).begin(), (a).end()
typedef vector<ll> vi;
typedef vector<vector<ll>> vvi;
typedef pair<ll, ll> pii;
const ll INF = 1LL << 58;
const ll MOD = 1000000007;
typedef vector<pii> vpii;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

typedef struct
{
    int me;
    vi child;
} node;

int ans[200010];
node tree[200010];

void solve(int id, vi v)
{
    auto itr = upper_bound(ALL(v), tree[id].me);
    int d = distance(v.begin(), itr);
    while (1)
    {
        v[d] = tree[id].me;
        if (v[d + 1] > tree[id].me)
        {
            break;
        }
        d++;
    }
    ans[id] = d;
    REP(i, tree[id].child.size())
    {
        if (ans[tree[id].child[i]] == -1)
            solve(tree[id].child[i], v);
    }
}

signed main()
{
    //  以降 cin の入力元が 'input.txt' になる
    //std::ifstream in("input.txt");
    //std::cin.rdbuf(in.rdbuf());
    int N;
    cin >> N;

    REP(i, N)
    {
        cin >> tree[i + 1].me;
    }
    REP(i, N - 1)
    {
        int a, b;
        cin >> a >> b;
        tree[a].child.push_back(b);
        tree[b].child.push_back(a);
    }
    REP(i, N)
    {
        ans[i + 1] = -1;
    }

    vi v;
    v.resize(200010);
    REP(i, 200010)
    {
        v[i] = INF;
    }
    v[0] = -1;

    solve(1, v);

    REP(i, N)
    {
        cout << ans[i + 1] << endl;
    }

    return 0;
}