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,INF=0x3f3f3f3f;
int n,a[N],dp[N],ans[N];
int u,v;
vector<int>e[N];
void dfs(int u,int fa){
    int pos=lower_bound(dp,dp+n,a[u])-dp;
    int x,y;
    x=dp[pos];
    dp[pos]=a[u];
    ans[u]=lower_bound(dp,dp+n,INF)-dp;
    for(int i=0;i<e[u].size();++i){
        int v=e[u][i];
        if(v==fa)continue;
        dfs(v,u);
    }
    dp[pos]=x;
}
int main(){
    memset(dp,INF,sizeof dp);
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<n;++i){
        scanf("%d%d",&u,&v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;++i){
        printf("%d\n",ans[i]);
    }
    return 0;
}
