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;

const int inf = 1000000037;
const int maxn = 200005;

int n, from, to;
int a[maxn], lis[maxn], ans[maxn];
vector < int > g[maxn];

void dfs (int u, int p = -1) {
	int j = lower_bound(lis, lis + n, a[u]) - lis;
	int w = lis[j];
	lis[j] = a[u], ans[u] = int(lower_bound(lis, lis + n + 1, inf) - lis) - 1;
	for (int v: g[u]) {
		if (v != p)
			dfs(v, u);
	}
	lis[j] = w;
}

signed main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	lis[0] = -inf;
	for (int i = 1; i <= n + 1; ++i)
		lis[i] = inf;
	for (int i = 1; i < n; ++i) {
		scanf("%d %d", &from, &to);
		g[from].push_back(to);
		g[to].push_back(from);
	}
	dfs(1);
	for (int i = 1; i <= n; ++i)
		printf("%d\n", ans[i]);
}
