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<iostream>
#include<vector>
#include<algorithm>
using namespace std;

vector<int> v[200000], a, ret, tmp;
int n;

void dfs(int par, int pos){
    int len = lower_bound(tmp.begin(), tmp.end(), a[pos])-tmp.begin();
    int bef = tmp[len];
    tmp[len] = a[pos];
    ret[pos] = lower_bound(tmp.begin(), tmp.end(), 1<<30)-tmp.begin();
    for(int next : v[pos]){
        if(next == par) continue;
        dfs(pos, next);
    }
    tmp[len] = bef;
}

int main(){
    cin >> n;
    a.resize(n);
    ret.resize(n);
    tmp.resize(n+1, 1<<30);
    for(int i = 0; i < n; i++)  cin >> a[i];
    for(int i = 1; i < n; i++){
        int x, y;
        cin >> x >> y;
        x--, y--;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    dfs(-1, 0);
    for(int i = 0; i < n; i++){
        cout << ret[i] << endl;
    }
    return 0;
}
