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>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define tp tuple<int,int,int>
#define mtp make_tuple
using namespace std;

const int N=2e5+5;
int n,a[N],res[N],m,e[N],f[N];
vector<int>adj[N];

void dfs(int u,int p=-1)
{
    int prem=m;
    f[u]=lower_bound(e+1,e+m+1,a[u])-e;
    if(f[u]>m) m++;
    int pref=e[f[u]];
    e[f[u]]=a[u];
    res[u]=m;
    for(auto&v:adj[u])
    {
        if(v==p) continue;
        dfs(v,u);
    }
    m=prem;
    e[f[u]]=pref;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1);
    for(int i=1;i<=n;i++) cout<<res[i]<<'\n';
}
