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>
#define rep(i,n) for (int i = 0; i < n; ++i)
using namespace std;
using ll = long long;
const int SIZE = 200000;

vector<int> G[SIZE];
vector<int> ans(SIZE), a(SIZE);
vector<int> L;
void dfs(int k = 0, int p = -1){
    int flag = -1;
    int itr = lower_bound(L.begin(), L.end(), a[k]) - L.begin();
    if (itr == L.size()){
        L.push_back(a[k]);
    }
    else {
        flag = L[itr];
        L[itr] = a[k];
    }
    ans[k] = L.size();
    for (auto &&i : G[k]){
        if (i == p)continue;
        dfs(i, k);
    }
    if (flag == -1)L.pop_back();
    else L[itr] = flag;
}
int main() {
    cin.tie(nullptr); ios::sync_with_stdio(false);
    int n; cin >> n;
    rep(i,n)cin >> a[i];
    rep(i,n-1){
        int u, v; cin >> u >> v;
        u--; v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs();
    rep(i,n)
        cout << ans[i] << endl;
}