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
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

YES
1 2 3
NO
YES
-4 -1 -3 2
YES
-3 -2 1 6 7 -8 -5 4

#include <algorithm>
#include <bitset>
#include <cassert>
#include <deque>
#include <iomanip>
#include <iostream>
#include <numeric>
#include <map>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <vector>

using std::cin, std::cout, std::endl;

void solve() {
  int N;
  cin >> N;
  std::vector<int> P(N);
  for (auto &El : P)
    cin >> El;

  using State = std::array<std::array<int, 2>, 2>; // [Count][IsNegated]
  const State INIT = State{std::array<int, 2>{0, 0}, {0, 0}};
  std::vector<State> DP(N, INIT);
  std::vector<State> From(N, INIT);
  DP[0] = State{std::array<int, 2>{1, 1}, {0, 0}};
  for (int Pos = 1; Pos < N; Pos++) {
    for (int Count = 0; Count < 2; Count++)
      for (int Neg = 0; Neg < 2; Neg++)
          for (int PrevNeg = 0; PrevNeg < 2; PrevNeg++) {
            int Cur = Neg ? -P[Pos] : P[Pos];
            int Prev = PrevNeg ? -P[Pos - 1] : P[Pos - 1];
            int PrevCount = Count - (Cur < Prev);
            if (PrevCount < 0)
              continue;
            if (DP[Pos - 1][PrevCount][PrevNeg]) {
              DP[Pos][Count][Neg] = 1;
              From[Pos][Count][Neg] = PrevCount * 2 + PrevNeg;
            }
          }
  }
  std::pair<int, int> Cur = {-1, -1};
  for (int Count = 0; Count < 2; Count++) {
    for (int Neg = 0; Neg < 2; Neg++) {
      if (DP[N - 1][Count][Neg])
        Cur = {Count, Neg};
    }
  }
  if (Cur.first == -1) {
    cout << "NO\n";
    return;
  }
  for (int I = N - 1; I >= 0; I--) {
    if (Cur.second)
      P[I] = -P[I];
    int Mask = From[I][Cur.first][Cur.second];
    Cur = {Mask >> 1, Mask & 1};
  }
  cout << "YES\n";
  for (auto El : P)
    cout << El << ' ';
  cout << '\n';
}

int main() {
  std::ios::sync_with_stdio(false);
  int T = 1;
  cin >> T;
  for (int TestCase = 1; TestCase <= T; TestCase++) {
    solve();
  }
  return 0;
}
