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 rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MX = 200005;
int n, a[MX], u, v, nw[MX], e, ans[MX];
vector<int> G[MX];
void rec(int t, int p){
	int* pos = lower_bound(nw, nw+e, a[t]);
	int pre = *pos;
	*pos = a[t];
	if(pos - nw == e) e++;
	ans[t] = e;
	for(int nx: G[t]) if(nx != p) rec(nx, t);
	*pos = pre;
	if(pre == 0) e--;
}
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%d", a+i);
	rep(i,n-1){
		scanf("%d%d", &v, &u);
		v--; u--;
		G[v].push_back(u);
		G[u].push_back(v);
	}
	rec(0, 0);
	rep(i,n) printf("%d\n", ans[i]);
}