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;

#define endl '\n'

int const N = 200020, inf = (int) 2e9;
int n, a[N], lis[N], ans[N];
vector<int> adj[N];

void dfs(int v = 0, int p = -1){
    int pos = (int) (lower_bound(lis, lis + N, a[v]) - lis);
    int ant = lis[pos];
    lis[pos] = a[v];
    ans[v] = (int) (lower_bound(lis, lis+N, inf) - lis) - 1;
    for (int f : adj[v]){
        if (f == p) continue;
        dfs(f, v);
    }
    lis[pos] = ant;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    lis[0] = -inf;
    for (int i=1; i<N; i++) lis[i] = inf;

    cin >> n;
    for (int i=0; i<n; i++) cin >> a[i];
    for (int i=0; i<n-1; i++){
        int u,v; cin >> u >> v; u--; v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    
    dfs();
    for (int i=0; i<n; i++) cout << ans[i] << endl;
}