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 pair<int,int> P;
const int N=200000,inf=1000000000;
vector<vector<int>>pas(N);
vector<int>dp(N,inf);
int lis[N],a[N];
stack<P>update;
void dfs(int x,int before){
    int l=lower_bound(dp.begin(),dp.end(),a[x])-dp.begin();
    update.push({l,dp[l]});
    dp[l]=a[x];
    lis[x]=lower_bound(dp.begin(),dp.end(),inf)-dp.begin();
    for(int i=0;i<pas[x].size();i++)if(pas[x][i]!=before)dfs(pas[x][i],x);
    P reset=update.top();
    update.pop();
    dp[reset.first]=reset.second;
    return;
}
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++)cin>>a[i];
    int v[n-1],u[n-1];
    for(int i=0;i<n-1;i++){
        cin>>v[i]>>u[i];
        pas[v[i]-1].push_back(u[i]-1);
        pas[u[i]-1].push_back(v[i]-1);
    }
    dfs(0,-1);
    for(int i=0;i<n;i++)cout<<lis[i]<<endl;
    return 0;
}