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;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 1000006, INF = 1000000009;
  int n;
  int a[N];
  int dp[N][2];
  int p[N][2];
  void minh(int& x, int& px, int y, int u) {
    if (y < x) {
      x = y;
      px = u;
    }
  }
  bool solv() {
    dp[1][0] = dp[1][1] = -INF;
    for (int i = 2; i <= n; ++i) dp[i][0] = dp[i][1] = INF;
    for (int i = 1; i < n; ++i) {
      if (a[i] < a[i + 1]) {
        minh(dp[i + 1][0], p[i + 1][0], dp[i][0], 0);
      }
      if (dp[i][0] < a[i + 1]) {
        minh(dp[i + 1][0], p[i + 1][0], a[i], 0);
      }
      if (dp[i][0] < -a[i + 1]) {
        minh(dp[i + 1][1], p[i + 1][1], a[i], 0);
      }
      if (-a[i] < a[i + 1]) {
        minh(dp[i + 1][0], p[i + 1][0], dp[i][1], 1);
      }
      if (-a[i] < -a[i + 1]) {
        minh(dp[i + 1][1], p[i + 1][1], dp[i][1], 1);
      }
      if (dp[i][1] < a[i + 1]) {
        minh(dp[i + 1][0], p[i + 1][0], -a[i], 1);
      }
      if (dp[i][1] < -a[i + 1]) {
        minh(dp[i + 1][1], p[i + 1][1], -a[i], 1);
      }
    }
    if (dp[n][0] != INF || dp[n][1] != INF) {
      int u;
      if (dp[n][0] != INF)
        u = 0;
      else
        u = 1;
      for (int i = n; i >= 1; --i) {
        if (u == 1) a[i] *= -1;
        u = p[i][u];
      }
      cout << "YES\n";
      for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
      cout << "\n";
      return true;
    }
    cout << "NO\n";
    return false;
  }
  bool solv0() {
    for (int x = 0; x < (1 << n); ++x) {
      for (int i = 1; i <= n; ++i) {
        if ((x & (1 << (i - 1)))) a[i] *= -1;
      }
      bool z = true;
      for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
          for (int k = j + 1; k <= n; ++k) {
            if (a[i] > a[j] && a[j] > a[k]) {
              z = false;
              break;
            }
          }
          if (!z) break;
        }
        if (!z) break;
      }
      if (z) {
        return true;
        for (int i = 1; i <= n; ++i) {
          if ((x & (1 << (i - 1)))) a[i] *= -1;
        }
        return true;
      }
      for (int i = 1; i <= n; ++i) {
        if ((x & (1 << (i - 1)))) a[i] *= -1;
      }
    }
    return false;
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 100;
    cin >> tt;
    while (tt--) {
      cin >> n;
      for (int i = 1; i <= n; ++i) cin >> a[i];
      solv();
    }
    return 0;
  }
  