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;
  inline int read() {
    char c = getchar();
    int x = 0;
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x;
    return x;
  }
  int n, p[1000005];
  int f[1000005][2], pre[1000005][2];
  void chkmin(int x, int y, int z, int o) {
    if (z < f[x][y]) f[x][y] = z, pre[x][y] = o;
  }
  int res[1000005];
  void work() {
    n = read();
    for (register int i = (1); i <= (n); ++i) p[i] = read();
    f[1][0] = -0x3f3f3f3f, f[1][1] = -0x3f3f3f3f;
    for (register int i = (2); i <= (n); ++i) {
      f[i][0] = f[i][1] = 0x3f3f3f3f;
      for (register int j = (0); j <= (1); ++j)
        for (register int k = (0); k <= (1); ++k) {
          int now = (j ? p[i] : -p[i]);
          int lst = (k ? p[i - 1] : -p[i - 1]);
          if (now > lst) chkmin(i, j, f[i - 1][k], k);
          if (now > f[i - 1][k]) chkmin(i, j, lst, k);
        }
    }
    if (f[n][0] > n && f[n][1] > n) {
      puts("NO");
      return;
    }
    puts("YES");
    int o = (f[n][1] <= n);
    res[n] = (o ? p[n] : -p[n]);
    for (register int i = (n - 1); i >= (1); --i) {
      o = pre[i + 1][o];
      res[i] = (o ? p[i] : -p[i]);
    }
    for (register int i = (1); i <= (n); ++i) printf("%d ", res[i]);
    puts("");
  }
  signed main() {
    int T = read();
    while (T--) work();
    return 0;
  }
  