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 p[1000005], dp[1000005][2][2], n;
  struct Node {
    int i, j, k;
  } last[1000005][2][2];
  void Min(int &a, int b, Node c, Node &d) {
    if (a > b) a = b, d = c;
  }
  void Main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
    for (int i = 0; i <= n; ++i)
      for (int j = 0; j < 2; ++j)
        for (int k = 0; k < 2; ++k) dp[i][j][k] = (int)1e9;
    dp[1][0][0] = -(int)1e9;
    dp[1][0][1] = -(int)1e9;
    for (int i = 1; i < n; ++i)
      for (int j = 0; j < 2; ++j)
        for (int k = 0; k < 2; ++k) {
          int a, b;
          if (j == 0) a = (k == 0 ? -p[i] : p[i]), b = dp[i][j][k];
          if (j == 1) a = dp[i][j][k], b = (k == 0 ? -p[i] : p[i]);
          if (p[i + 1] > a)
            Min(dp[i + 1][0][1], b, (Node){i, j, k}, last[i + 1][0][1]);
          else if (p[i + 1] > b)
            Min(dp[i + 1][1][1], a, (Node){i, j, k}, last[i + 1][1][1]);
          if (-p[i + 1] > a)
            Min(dp[i + 1][0][0], b, (Node){i, j, k}, last[i + 1][0][0]);
          else if (-p[i + 1] > b)
            Min(dp[i + 1][1][0], a, (Node){i, j, k}, last[i + 1][1][0]);
        }
    int nj = -1, nk = -1;
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k)
        if (dp[n][j][k] < (int)1e9) {
          nj = j;
          nk = k;
        }
    if (nj == -1) {
      puts("NO");
      return;
    }
    for (int i = n; i >= 1; --i) {
      if (!nk) p[i] = -p[i];
      int aj = last[i][nj][nk].j, ak = last[i][nj][nk].k;
      nj = aj;
      nk = ak;
    }
    puts("YES");
    for (int i = 1; i <= n; ++i) printf("%d%c", p[i], (i == n ? '\n' : ' '));
  }
  int main() {
    int T;
    scanf("%d", &T);
    for (; T--;) Main();
  }
  