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: #include <bits/stdc++.h>
  using namespace std;
  int T, n, p[1000005], q[1000005];
  int f[1000005][2][2], g[1000005][2][2];
  void sol() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
    for (int i = 1; i <= n; i++)
      f[i][0][0] = f[i][0][1] = f[i][1][0] = f[i][1][1] = n + 1;
    f[1][0][0] = f[1][1][0] = -n - 1;
    for (int i = 1; i < n; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++)
          if (f[i][j][k] <= n) {
            int x = j ? p[i] : -p[i], y = f[i][j][k];
            if (k) swap(x, y);
            for (int l = 0; l < 2; l++) {
              int z = l ? p[i + 1] : -p[i + 1];
              if (z < y) continue;
              if (z < x) {
                if (f[i + 1][l][1] > x) {
                  f[i + 1][l][1] = x;
                  g[i + 1][l][1] = j << 1 | k;
                }
              } else {
                if (f[i + 1][l][0] > y) {
                  f[i + 1][l][0] = y;
                  g[i + 1][l][0] = j << 1 | k;
                }
              }
            }
          }
    int t1 = -1, t2 = -1;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        if (f[n][i][j] <= n) t1 = i, t2 = j;
    if (t1 < 0) {
      puts("NO");
      return;
    }
    puts("YES");
    for (int i = n; i; i--) {
      q[i] = t1 ? p[i] : -p[i];
      if (i == 1) break;
      int msk = g[i][t1][t2];
      t1 = msk >> 1, t2 = msk & 1;
    }
    for (int i = 1; i <= n; i++) printf("%d ", q[i]);
    puts("");
  }
  int main() {
    scanf("%d", &T);
    while (T--) sol();
  }
  