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;
  const int N = 1e6 + 5;
  int n, p[N], dp[N][2][2];
  array<int, 3> pre[N][2][2];
  void reset(int a[][2]) {
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) a[i][j] = INT_MAX;
  }
  inline bool Min(int &x, int y) {
    if (x > y) {
      x = y;
      return true;
    }
    return false;
  }
  void echo(int a, int b, int c) {
    if (!a) return;
    echo(pre[a][b][c][0], pre[a][b][c][1], pre[a][b][c][2]);
    printf("%d%c", (c ? -p[a] : p[a]), " \n"[a == n]);
  }
  void solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", p + i);
    reset(dp[1]);
    dp[1][0][0] = -n;
    dp[1][0][1] = -n;
    for (int i = 2; i <= n; i++) {
      reset(dp[i]);
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++) {
          if (dp[i - 1][j][k] == INT_MAX) continue;
          int x = p[i - 1], y = dp[i - 1][j][k];
          if (k) x = -x;
          if (j) swap(x, y);
          if (p[i] >= x) {
            if (Min(dp[i][0][0], y)) {
              pre[i][0][0] = {i - 1, j, k};
            }
          } else if (p[i] >= y) {
            if (Min(dp[i][1][0], x)) {
              pre[i][1][0] = {i - 1, j, k};
            }
          }
          if (-p[i] >= x) {
            if (Min(dp[i][0][1], y)) {
              pre[i][0][1] = {i - 1, j, k};
            }
          } else if (-p[i] >= y) {
            if (Min(dp[i][1][1], x)) {
              pre[i][1][1] = {i - 1, j, k};
            }
          }
        }
    }
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) {
        if (dp[n][i][j] < INT_MAX) {
          puts("YES");
          echo(n, i, j);
          return;
        }
      }
    puts("NO");
  }
  int main() {
    int _;
    scanf("%d", &_);
    while (_--) solve();
    return 0;
  }
  