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;
typedef long long ll;
const int sz = 2e5+5;

int a[sz], ans[sz];
vector <int> g[sz];
multiset <int> mst;

void dfs(int u = 1, int p = 0)
{
    mst.insert(a[u]);
    auto it = mst.lower_bound(a[u]);
    it++;
    int x = 0;
    if(it != mst.end()) x = *it, mst.erase(it);

    for(int v : g[u]) if(v ^ p) dfs(v, u);

    ans[u] = mst.size();
    if(x) mst.insert(x);
    mst.erase(mst.find(a[u]));
}

int main()
{
    int n;
    cin >> n;
    for(int i=1; i<=n; i++) scanf("%d", &a[i]);

    for(int i=1; i<n; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }

    dfs();

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