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 rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
using P = pair<int,int>;
vector<ll> a,ans;
vector<ll> p;
vector<int> ki[220000];
const ll INF = 100100100100100100LL;
void dfs(int now,int last = -1){
    int ind = lower_bound(p.begin(),p.end(),a[now])- p.begin();
    ll old = p[ind];
    
    p[ind] = a[now];
    ans[now] = lower_bound(p.begin(),p.end(),INF)-p.begin() -1;
    
    for(auto next:ki[now]){
        if(next == last)continue;

        dfs(next,now);
    }
    p[ind] = old;
}
int main()
{
    int n;cin >> n;
    a.resize(n);
    ans.resize(n);
    p = vector<ll>(n+1,INF);
    p[0] = -INF;
    rep(i,n)cin >> a[i];
    rep(i,n-1){
        int a,b;
        cin >> a >> b;
        a--;b--;
        ki[a].push_back(b);
        ki[b].push_back(a);
    }
    dfs(0);
    rep(i,n)cout << ans[i] << endl;
    return 0;
}