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, inf = 1e9;
  int n, a[N], f[N][2][2], ans[N];
  void clean(int x) {
    for (int i = (0); i <= (1); ++i)
      for (int j = (0); j <= (1); ++j) f[x][i][j] = inf;
  }
  int fx[N][2][2], fy[N][2][2];
  void upd(int i, int j, int k, int px, int py, int x) {
    if (f[i][j][k] > x) f[i][j][k] = x, fx[i][j][k] = px, fy[i][j][k] = py;
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T, x, y, px, py, tx, ty;
    cin >> T;
    while (T--) {
      cin >> n;
      for (int i = (1); i <= (n); ++i) cin >> a[i];
      clean(1);
      f[1][1][0] = f[1][1][1] = -inf;
      for (int i = (2); i <= (n); ++i) {
        clean(i);
        for (int k1 = (0); k1 <= (1); ++k1)
          for (int k2 = (0); k2 <= (1); ++k2) {
            x = (k1 ? -a[i - 1] : a[i - 1]);
            y = (k2 ? -a[i] : a[i]);
            for (int j1 = (0); j1 <= (1); ++j1) {
              if (f[i - 1][j1][k1] == inf) continue;
              if (x < y) {
                if (j1 == 0) {
                  if (f[i - 1][j1][k1] > y)
                    upd(i, 0, k2, j1, k1, f[i - 1][j1][k1]);
                  else
                    upd(i, 1, k2, j1, k1, x);
                } else
                  upd(i, 1, k2, j1, k1, f[i - 1][j1][k1]);
              } else {
                if (j1 && y > f[i - 1][j1][k1]) upd(i, 0, k2, j1, k1, x);
              }
            }
          }
      }
      px = -1;
      py = -1;
      for (int j = (0); j <= (1); ++j)
        for (int k = (0); k <= (1); ++k)
          if (f[n][j][k] != inf) {
            px = j;
            py = k;
            break;
          }
      if (px == -1) {
        cout << "NO" << '\n';
        continue;
      }
      cout << "YES" << '\n';
      for (int i = (n); i >= (1); --i) {
        ans[i] = (py ? -a[i] : a[i]);
        tx = fx[i][px][py];
        ty = fy[i][px][py];
        px = tx;
        py = ty;
      }
      for (int i = (1); i <= (n); ++i) cout << ans[i] << ' ';
      cout << '\n';
    }
    return 0;
  }
  