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 N = 2e5+10;

int n, a[N];
vector<int> to[N];

int b[N], ans[N];

void dfs(int u, int f) {
	int pos = -1, val = 0;
	pos = lower_bound(b, b+n+1, a[u])-b;
	val = b[pos];
	b[pos] = min(b[pos], a[u]);
	for (int v: to[u]) {
		if (v == f) continue;
		dfs(v, u);
	}
	ans[u] = lower_bound(b, b+n+1, 0x3f3f3f3f)-b-1;
	b[pos] = val;
}

int main() {
	scanf("%d", &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);
		to[u].push_back(v);
		to[v].push_back(u);
	}
	memset(b, 0x3f, sizeof(b));
	b[0] = 0;
	dfs(1, 1);
	for (int i = 1; i <= n; ++i) {
		printf("%d\n", ans[i]);
	}
}
