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 ll;
const ll INF = 1LL<<60;
int N;
vector<int> A;
vector<vector<int>> G;
vector<ll> dp;
vector<int> res;
int ma;
void f(int p, int q) {
	ll x = A[p];
	auto y = (int)(lower_bound(dp.begin(), dp.end(), x) - dp.begin());
	auto tmp_ma = ma;
	ma = max(ma, y);
	res[p] = ma;
	auto tmp = dp[y];
	dp[y] = x;
	for (int v : G[p]) if (v != q) {
		f(v, p);
	}
	dp[y] = tmp;
	ma = tmp_ma;
}
int main() {
	cin >> N;
	A.resize(N);
	for (auto& x : A) cin >> x;
	G.resize(N);
	for (int i = 0; i < N-1; ++ i) {
		int u, v;
		cin >> u >> v;
		-- u; -- v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	res.resize(N);
	dp.resize(N+10, INF);
	dp[0] = -INF;
	f(0, -1);
	for (auto x : res) cout << x << endl;
}
