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.

Example

Input

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


Output

1
2
3
3
4
4
5
2
2
3
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll N;
vector<vector<ll>> G;
ll A[200000];
ll L[200000];
vector<ll> v;
ll INF = 1e15;

void dfs(ll n){
    ll ind = lower_bound(v.begin(),v.end(),A[n]) - v.begin();
    ll backup = v[ind];
    v[ind] = A[n];
    L[n] = lower_bound(v.begin(),v.end(),INF) - v.begin();
    for(auto &e : G[n]){
        if(L[e] != 0) continue;
        dfs(e);
    }
    v[ind] = backup;
}

int main(){
    cin >> N;
    G.resize(N);
    v.resize(N);
    for(int i=0;i<N;i++) cin >> A[i];
    for(int i=0;i<N-1;i++){
        ll a,b;
        cin >> a >> b;
        G[a-1].push_back(b-1);
        G[b-1].push_back(a-1);
    }
    for(int i=0;i<N;i++) v[i] = INF;
    dfs(0);
    for(int i=0;i<N;i++) cout << L[i] << endl;
    return 0;
}