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 = 1000000, INF = 0x3f3f3f3f;
  int aa[N], dp0[N], pp0[N], dp1[N], pp1[N], *dp;
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        aa[i] = a;
      }
      fill(dp0, dp0 + n, INF);
      fill(dp1, dp1 + n, INF);
      dp0[0] = dp1[0] = -INF;
      for (int i = 0; i + 1 < n; i++) {
        if (dp0[i] < aa[i + 1] && dp0[i + 1] > aa[i])
          dp0[i + 1] = aa[i], pp0[i + 1] = 1;
        if (dp1[i] < aa[i + 1] && dp0[i + 1] > -aa[i])
          dp0[i + 1] = -aa[i], pp0[i + 1] = -1;
        if (aa[i] < aa[i + 1] && dp0[i + 1] > dp0[i])
          dp0[i + 1] = dp0[i], pp0[i + 1] = 1;
        if (-aa[i] < aa[i + 1] && dp0[i + 1] > dp1[i])
          dp0[i + 1] = dp1[i], pp0[i + 1] = -1;
        if (dp0[i] < -aa[i + 1] && dp1[i + 1] > aa[i])
          dp1[i + 1] = aa[i], pp1[i + 1] = 1;
        if (dp1[i] < -aa[i + 1] && dp1[i + 1] > -aa[i])
          dp1[i + 1] = -aa[i], pp1[i + 1] = -1;
        if (-aa[i] < -aa[i + 1] && dp1[i + 1] > dp1[i])
          dp1[i + 1] = dp1[i], pp1[i + 1] = -1;
      }
      if (dp0[n - 1] != INF || dp1[n - 1] != INF) {
        int p = dp0[n - 1] != INF ? 1 : -1;
        cout << "YES\n";
        for (int i = n - 1; i >= 0; i--) {
          aa[i] *= p;
          p = p == 1 ? pp0[i] : pp1[i];
        }
        for (int i = 0; i < n; i++) cout << aa[i] << ' ';
        cout << '\n';
      } else
        cout << "NO\n";
    }
    return 0;
  }
  