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>
#include <functional>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> a(N);
    for (int &t: a)
        cin>>t;
    vector<vector<int>> E(N);
    for (int i=0; i<N-1; i++)
    {
        int u, v;
        cin>>u>>v;
        E[u-1].push_back(v-1);
        E[v-1].push_back(u-1);
    }

    int oo = 1111111111;
    vector<int> T(N, oo);
    vector<int> ans(N);
    function<void(int, int)> f = [&](int c, int p)
    {
        int x = int(lower_bound(T.begin(), T.end(), a[c])-T.begin());
        int old = T[x];
        T[x] = a[c];
        ans[c] = int(lower_bound(T.begin(), T.end(), oo)-T.begin());
        for (int e: E[c])
            if (e!=p)
                f(e, c);
        T[x] = old;
    };
    f(0, -1);
    for (int x: ans)
        cout<<x<<endl;
}
