description: 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
solution: // coached by rainboy
  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();
  	}
  }
  