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;
#define ll long long int
int n;
vector<int> a;
vector<int> adj[200007];

vector<int> vv,ans;
multiset<int> s;
void dfs(int u, int p){
    s.insert(a[u]);
	int del=-1;
	auto it=s.lower_bound(a[u]);
	if(++it!=s.end()) del=*it,s.erase(it);
	ans[u]=s.size();
    for(auto v:adj[u]){
        if(v!=p){
            dfs(v,u);
        }
    }
    if(del!=-1) s.insert(del);
	s.erase(s.find(a[u]));
}

int main()
{
    cin >> n;
    a.resize(n+1);
    ans.resize(n+1);
    for(int i=1;i<=n;i++){
        cin >> a[i];
    }
    for(int i=0;i<n-1;i++){
        int u,v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++){
        cout << ans[i] << endl;
     }
}
