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;
const int N=2e5+10;
int a[N];
const int MAX=1e9+10;
vector <int> g[N];
int f[N];
int n;
int ans[N];
void dfs(int fa,int now){
    int pos=lower_bound(f,f+n+1,a[now])-f;
    int tmp=f[pos];
    f[pos]=min(a[now],f[pos]);
    for(int i=0;i<g[now].size();i++){
        if(g[now][i]==fa) continue;
        dfs(now,g[now][i]);
    }
    ans[now]=lower_bound(f,f+n+1,MAX)-f-1;
    f[pos]=tmp;
}
int main()
{
    //cout<<0x3f;
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=n-1;i++){
        int x,y;
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    memset(f,0x3f,sizeof(f));
    f[0]=0;
    dfs(1,1);
    for(int i=1;i<=n;i++) cout<<ans[i]<<endl;
}
