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 long long read() {
    long long x = 0, f = 1;
    char c = getchar();
    while ((c < '0' || c > '9') && (c != '-')) c = getchar();
    if (c == '-') f = -1, c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
  }
  const int N = 1e6 + 10;
  int n, a[N][2], dp[N][2][2], ans[N];
  pair<int, int> pre[N][2][2];
  inline void solve() {
    n = read();
    for (int i = (1); i <= (n); i++) a[i][0] = read(), a[i][1] = -a[i][0];
    if (n == 1) return puts("YES\n1\n"), void(0);
    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] = n + 1, pre[i][j][k] = make_pair(-1, -1);
    for (int j = (0); j <= (1); j++) dp[1][j][0] = -n - 1;
    for (int i = (1); i < (n); i++)
      for (int j = (0); j <= (1); j++)
        for (int k = (0); k <= (1); k++)
          if (dp[i][j][k] < n + 1) {
            for (int l = (0); l <= (1); l++) {
              int x = 0, y = 0;
              if (k == 0)
                x = a[i][j], y = dp[i][j][k];
              else
                y = a[i][j], x = dp[i][j][k];
              int z = a[i + 1][l];
              if (z < y) continue;
              if (z > x) {
                if (y < dp[i + 1][l][0])
                  dp[i + 1][l][0] = y, pre[i + 1][l][0] = make_pair(j, k);
              } else {
                if (x < dp[i + 1][l][1])
                  dp[i + 1][l][1] = x, pre[i + 1][l][1] = make_pair(j, k);
              }
            }
          }
    int x = -1, y = -1;
    for (int j = (0); j <= (1); j++)
      for (int k = (0); k <= (1); k++)
        if (pre[n][j][k].first != -1) x = j, y = k;
    if (x == -1) return puts("NO"), void(0);
    for (int i = (n); i >= (1); i--) {
      ans[i] = a[i][x];
      auto tmp = pre[i][x][y];
      x = tmp.first, y = tmp.second;
    }
    puts("YES");
    for (int i = (1); i <= (n); i++) printf("%d ", ans[i]);
    puts("");
  }
  int main() {
    int T = read();
    while (T--) solve();
  }
  