You are given a permutation p consisting of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).

Let's call an array a bipartite if the following undirected graph is bipartite:

  * the graph consists of n vertices; 
  * two vertices i and j are connected by an edge if i < j and a_i > a_j. 



Your task is to find a bipartite array of integers a of size n, such that a_i = p_i or a_i = -p_i, or report that no such array exists. If there are multiple answers, print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^6) — the size of the permutation.

The second line contains n integers p_1, p_2, ..., p_n.

The sum of n over all test cases doesn't exceed 10^6.

Output

For each test case, print the answer in the following format. If such an array a does not exist, print "NO" in a single line. Otherwise, print "YES" in the first line and n integers — array a in the second line.

Example

Input


4
3
1 2 3
6
1 3 2 6 5 4
4
4 1 3 2
8
3 2 1 6 7 8 5 4


Output


YES
1 2 3
NO
YES
-4 -1 -3 -2
YES
-3 -2 1 6 7 -8 -5 -4
4
3
1 2 3
6
1 3 2 6 5 4
4
4 1 3 2
8
3 2 1 6 7 8 5 4

YES
1 2 3
NO
YES
-4 -1 -3 2
YES
-3 -2 1 6 7 -8 -5 4

import java.io.*;
import java.util.*;

public class Solution extends PrintWriter {

	static final int INF = 1000001;
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] aa = new int[n + 1];
			for (int i = 0; i < n; i++)
				aa[i] = sc.nextInt();
			aa[n] = n + 1;
			int[] dp = new int[n + 1];
			int[] dq = new int[n + 1];
			dp[0] = dq[0] = -INF;
			for (int i = 1; i <= n; i++) {
				int p = INF, q = INF;
				int a = aa[i - 1], b = aa[i];
				int x = dp[i - 1];
				int l = Math.min(a, x);
				int r = Math.max(a, x);
				if (b > r)
					p = Math.min(p, l);
				else if (b > l)
					p = Math.min(p, r);
				if (-b > r)
					q = Math.min(q, l);
				else if (-b > l)
					q = Math.min(q, r);
				x = dq[i - 1];
				l = Math.min(-a, x);
				r = Math.max(-a, x);
				if (b > r)
					p = Math.min(p, l);
				else if (b > l)
					p = Math.min(p, r);
				if (-b > r)
					q = Math.min(q, l);
				else if (-b > l)
					q = Math.min(q, r);
				dp[i] = p; dq[i] = q;
			}
			if (dp[n] == INF)
				println("NO");
			else {
				println("YES");
				for (int i = n - 1; i >= 0; i--) {
					int a = aa[i], b = aa[i + 1];
					int x = b > 0 ? dp[i + 1] : dq[i + 1];
					if (a < b && dp[i] == x)
						continue;
					if (dp[i] < b && a == x)
						continue;
					aa[i] = -aa[i];
				}
				for (int i = 0; i < n; i++)
					print(aa[i] + " ");
				println();
			}
		}
	}
	Solution() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		Solution o = new Solution(); o.main(); o.flush();
	}
}
