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>

#ifdef NON_SUBMIT
#define TEST(n) (n)
#define tout cerr
#else
#define TEST(n) ((void)0)
#define tout cin
#endif

using namespace std;

vector<int> adj[200000];
int N, x[200000], A[200000], ans[200000];
bool vis[200000];

void dfs(int c)
{
	vis[c]=true;
	ans[c]=lower_bound(x,x+N,0x7f7f7f7f)-x;
	for(auto n: adj[c]) if(!vis[n]) {
		int r=lower_bound(x,x+N,A[n])-x, v=A[n];
		swap(x[r],v);
		dfs(n);
		swap(x[r],v);
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	TEST(freopen("input.txt","r",stdin));
	TEST(freopen("output.txt","w",stdout));
	TEST(freopen("debug.txt","w",stderr));
	cin>>N; memset(x,0x7f,sizeof(x));
	for(int i=0;i<N;i++) cin>>A[i];
	for(int i=1;i<N;i++) {
		int a, b;
		cin>>a>>b;
		adj[--a].push_back(--b);
		adj[b].push_back(a);
	}
	x[0]=A[0];
	dfs(0);
	for(int i=0;i<N;i++) cout<<ans[i]<<'\n';
	return 0;
}