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() {
    int ans = 0, f = 1;
    char c = getchar();
    while (c > '9' || c < '0') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      ans = (ans << 1) + (ans << 3) + c - '0';
      c = getchar();
    }
    return ans * f;
  }
  inline void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x / 10) write(x / 10);
    putchar((char)(x % 10) + '0');
  }
  template <typename T>
  inline T Abs(T a) {
    return a > 0 ? a : -a;
  };
  template <typename T, typename TT>
  inline T Min(T a, TT b) {
    return a < b ? a : b;
  }
  template <typename T, typename TT>
  inline T Max(T a, TT b) {
    return a < b ? b : a;
  }
  const int N = 1e6 + 5;
  int t, n, a[N], dp[N][2][2], p1[N][2][2], p2[N][2][2], p3[N][2][2];
  inline void update(int x, int y, int z, int v, int Y, int Z) {
    if (dp[x][y][z] > v) {
      dp[x][y][z] = v;
      p2[x][y][z] = Y;
      p3[x][y][z] = Z;
    }
  }
  void print(int x, int y, int z) {
    if (!x) return;
    int X, Y;
    if (y == 0) X = (z ? a[x] : -a[x]), Y = dp[x][y][z];
    if (y == 1) Y = (z ? a[x] : -a[x]), X = dp[x][y][z];
    print(x - 1, p2[x][y][z], p3[x][y][z]);
    if (Abs(X) == a[x]) write(X);
    if (Abs(Y) == a[x]) write(Y);
    putchar(' ');
  }
  signed main() {
    t = read();
    while (t--) {
      n = read();
      for (int i = 1; i <= n; ++i) a[i] = read();
      for (int i = 1; i <= n; ++i)
        for (int j = 0; j <= 1; ++j)
          for (int k = 0; k <= 1; ++k) dp[i][j][k] = 1e9;
      dp[1][0][0] = -1e9;
      dp[1][0][1] = -1e9;
      for (int i = 1; i <= n - 1; ++i)
        for (int j = 0; j <= 1; ++j)
          for (int k = 0; k <= 1; ++k)
            if (dp[i][j][k] != 1e9) {
              int x, y;
              if (j == 0) x = (k ? a[i] : -a[i]), y = dp[i][j][k];
              if (j == 1) y = (k ? a[i] : -a[i]), x = dp[i][j][k];
              if (a[i + 1] > y) {
                if (a[i + 1] > x)
                  update(i + 1, 0, 1, y, j, k);
                else
                  update(i + 1, 1, 1, x, j, k);
              }
              if (-a[i + 1] > y) {
                if (-a[i + 1] > x)
                  update(i + 1, 0, 0, y, j, k);
                else
                  update(i + 1, 1, 0, x, j, k);
              }
            }
      int flag = 1;
      for (int i = 0; i <= 1 && flag; ++i)
        for (int j = 0; j <= 1 && flag; ++j)
          if (dp[n][i][j] != 1e9) {
            puts("YES");
            print(n, i, j);
            puts("");
            flag = 0;
          }
      if (flag) puts("NO");
    }
    return 0;
  }
  