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 <iostream>
#include <vector>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int n;
int a[200000];
vector<int> et[200000];

int INF = 1e+10;
int dp[200001];	//dp[length] = min_value
int ans[200001];
void dfs(int p, int v, int len) {
	int pos = lower_bound(dp, dp + len, a[v]) - dp;
	int tmp = dp[pos];
	int nlen = len;
	if (pos == len) { nlen = len + 1; }
	dp[pos] = a[v];
	
	ans[v] = nlen;
	int i;
	rep(i, et[v].size()) {
		int nv = et[v][i];
		if (nv == p) continue;
		dfs(v, nv, nlen);
	}
	dp[pos] = tmp;
}

signed main() {
	int i;
	
	cin >> n;
	rep(i, n) cin >> a[i];
	rep(i, n - 1) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		et[u].push_back(v);
		et[v].push_back(u);
	}
	
	rep(i, n + 1) dp[i] = INF;
	dfs(-1, 0, 0);
	rep(i, n) cout << ans[i]  << endl;
	return 0;
}
