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;

int n;
vector <int> a ,ans;
vector <vector<int>> adj;

set <int> lis;
void dfs(int u ,int p){
    int bef = -1;
    auto it = lis.lower_bound(a[u]);
    if(it != lis.end())
        lis.erase(it) ,bef = *it;
    lis.insert(a[u]);
    ans[u] = lis.size();
    for(int&v : adj[u]) if(v != p)
        dfs(v ,u);
    if(lis.count(a[u]))
        lis.erase(a[u]);
    if(~bef)
        lis.insert(bef);
}

int main()
{
    scanf("%d",&n);
    a.resize(n+1);
    ans.resize(n+1);
    adj.resize(n+1);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    for(int a,b,i=1; i<n; i++){
        scanf("%d%d",&a,&b);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    dfs(1 ,-1);

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