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>
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  constexpr long long mod = 1000000007;
  const long long INF = mod * mod;
  const double eps = 1e-4;
  const double pi = acosl(-1.0);
  template <typename T>
  void chmin(T& a, T b) {
    a = min(a, b);
  }
  template <typename T>
  void chmax(T& a, T b) {
    a = max(a, b);
  }
  template <typename T>
  void cinarray(vector<T>& v) {
    for (int i = 0; i < v.size(); i++) cin >> v[i];
  }
  template <typename T>
  void coutarray(vector<T>& v) {
    for (int i = 0; i < v.size(); i++) {
      if (i > 0) cout << " ";
      cout << v[i];
    }
    cout << "\n";
  }
  long long mod_pow(long long x, long long n, long long m = mod) {
    if (n < 0) {
      long long res = mod_pow(x, -n, m);
      return mod_pow(res, m - 2, m);
    }
    if (abs(x) >= m) x %= m;
    if (x < 0) x += m;
    if (x == 0) return 0;
    long long res = 1;
    while (n) {
      if (n & 1) res = res * x % m;
      x = x * x % m;
      n >>= 1;
    }
    return res;
  }
  struct modint {
    int n;
    modint() : n(0) { ; }
    modint(long long m) {
      if (m < 0 || mod <= m) {
        m %= mod;
        if (m < 0) m += mod;
      }
      n = m;
    }
    operator int() { return n; }
  };
  bool operator==(modint a, modint b) { return a.n == b.n; }
  modint operator+=(modint& a, modint b) {
    a.n += b.n;
    if (a.n >= mod) a.n -= mod;
    return a;
  }
  modint operator-=(modint& a, modint b) {
    a.n -= b.n;
    if (a.n < 0) a.n += mod;
    return a;
  }
  modint operator*=(modint& a, modint b) {
    a.n = ((long long)a.n * b.n) % mod;
    return a;
  }
  modint operator+(modint a, modint b) { return a += b; }
  modint operator-(modint a, modint b) { return a -= b; }
  modint operator*(modint a, modint b) { return a *= b; }
  modint operator^(modint a, long long n) {
    if (n == 0) return modint(1);
    modint res = (a * a) ^ (n / 2);
    if (n % 2) res = res * a;
    return res;
  }
  long long inv(long long a, long long p) {
    return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
  }
  modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
  modint operator/=(modint& a, modint b) {
    a = a / b;
    return a;
  }
  const int max_n = 1 << 20;
  modint fact[max_n], factinv[max_n];
  void init_f() {
    fact[0] = modint(1);
    for (int i = 0; i < max_n - 1; i++) {
      fact[i + 1] = fact[i] * modint(i + 1);
    }
    factinv[max_n - 1] = modint(1) / fact[max_n - 1];
    for (int i = max_n - 2; i >= 0; i--) {
      factinv[i] = factinv[i + 1] * modint(i + 1);
    }
  }
  modint comb(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[b] * factinv[a - b];
  }
  modint combP(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[a - b];
  }
  long long gcd(long long a, long long b) {
    a = abs(a);
    b = abs(b);
    if (a < b) swap(a, b);
    while (b) {
      long long r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  int dx[4] = {1, 0, -1, 0};
  int dy[4] = {0, 1, 0, -1};
  int dp[1 << 20][2];
  int pre[1 << 20][2];
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n + 1; i++) {
      for (int j = 0; j < 2; j++) {
        dp[i][j] = mod;
      }
    }
    dp[1][0] = dp[1][1] = -mod;
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++)
          for (int l = 0; l < 2; l++) {
            pair<int, int> p;
            p.first = a[i - 1];
            if (j == 0) p.first *= -1;
            p.second = dp[i][j];
            if (k) swap(p.first, p.second);
            int val = a[i];
            if (l == 0) val *= -1;
            if (p.first < val) {
              if (dp[i + 1][l] > p.second) {
                dp[i + 1][l] = p.second;
                pre[i + 1][l] = j;
              }
            }
          }
    }
    if (dp[n][0] == mod && dp[n][1] == mod) {
      cout << "NO\n";
      return;
    }
    vector<int> ans = a;
    int cur = 0;
    if (dp[n][0] == mod) cur = 1;
    for (int i = n - 1; i >= 0; i--) {
      if (cur == 0) ans[i] *= -1;
      cur = pre[i + 1][cur];
    }
    cout << "YES\n";
    coutarray(ans);
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) solve();
    return 0;
  }
  