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;
  char buf[1 << 21], *p1 = buf, *p2 = buf;
  inline long long rd() {
    char c =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
    long long x = 0;
    bool f = 0;
    for (; !isdigit(c);
         c = (p1 == p2 &&
                      (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                  ? EOF
                  : *p1++))
      f ^= !(c ^ 45);
    for (; isdigit(c);
         c = (p1 == p2 &&
                      (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                  ? EOF
                  : *p1++))
      x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x;
    return x;
  }
  int i, j, k, n, m, t, it, a[1005000], fa[10005000];
  bool op[10050000];
  struct SB {
    int id, x, y;
    bool operator<(const SB a) const { return (x == a.x) ? y < a.y : x < a.x; }
  };
  vector<SB> v1, v2;
  int main() {
    ios::sync_with_stdio(0);
    t = rd();
    while (t--) {
      n = rd();
      it = 0;
      for (i = 1; i <= n; i++) a[i] = rd();
      v1.push_back({0, -11451419, -11451419});
      for (i = 1; i <= n; i++) {
        v2.clear();
        for (auto &[id, x, y] : v1) {
          if (x < -a[i]) {
            fa[++it] = id;
            op[it] = 1;
            v2.push_back({it, -a[i], y});
            goto aaa;
          }
          if (y < -a[i]) {
            fa[++it] = id;
            op[it] = 1;
            v2.push_back({it, x, -a[i]});
            goto aaa;
          }
        aaa:;
          if (x < a[i]) {
            fa[++it] = id;
            op[it] = 0;
            v2.push_back({it, a[i], y});
            goto bbb;
          }
          if (y < a[i]) {
            fa[++it] = id;
            op[it] = 0;
            v2.push_back({it, x, a[i]});
            goto bbb;
          }
        bbb:;
        }
        if (v2.empty()) {
          puts("NO");
          goto ccc;
        }
        v1.clear();
        k = 1e9;
        sort(v2.begin(), v2.end());
        for (auto &i : v2) {
          if (i.y < k) {
            k = i.y;
            v1.push_back(i);
          }
        }
      }
      i = (*v1.begin()).id;
      j = n;
      while (i) {
        if (op[i]) a[j] *= -1;
        j--;
        i = fa[i];
      }
      puts("YES");
      for (i = 1; i <= n; i++) printf("%d ", a[i]);
      puts("");
    ccc:;
    }
  }
  