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 <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN=2e5+10;
const int INF=1e9+10;
int N;
int A[MAXN];
int ans[MAXN];
int dp[MAXN];
vector<int> graph[MAXN];

void DFS(int cur,int parent,int max_len){
    int *pos=lower_bound(dp+1,dp+1+max_len,A[cur]);
    auto temp=*pos;
    *pos=A[cur];
    auto len=pos-dp;
    // printf("len=%d\n",len);
    auto new_max=max(max_len,(int)len);
    ans[cur]=new_max;
    for(auto adj:graph[cur]){
        if (adj==parent) continue;
        DFS(adj,cur,new_max);
    }

    *pos=temp;
}

int main(){
    scanf("%d",&N);
    fill(dp+1,dp+3+N,INF);
    for(int i=1;i<=N;i++) scanf("%d",A+i);
    for(int i=0;i<N-1;i++){
        int s,e;
        scanf("%d%d",&s,&e);
        graph[s].push_back(e);
        graph[e].push_back(s);
    }

    DFS(1,1,0);
    for(int i=1;i<=N;i++) printf("%d\n",ans[i]);

    return 0;
}