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 int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int n;
VI a(214514);
VVI edge(214514, VI(0));
VI ans(214514);
VI dp(214514,LINF);

void dfs(int v, int p) {
	int itr=lower_bound(ALL(dp), a[v])-dp.begin();
	ll z=dp[itr];
	dp[itr]=a[v];
	ans[v]=lower_bound(ALL(dp), INF)-dp.begin();
	for (auto to : edge[v])
		if (to != p) dfs(to, v);
	dp[itr]=z;
}

int main() {
	cin >> n;
	REP(i,n) cin >> a[i];
	int u, v;
	REP(i,n-1) {
		cin >> u >> v;
		u--; v--;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	dfs(0,-1);
	REP(i,n)
		cout << ans[i] << endl;
	return 0;
}