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 <bits/stdc++.h>
using std::cin, std::cout, std::endl;
void solve() {
  int N;
  cin >> N;
  std::vector<int> P(N);
  for (auto &El : P) cin >> El;
  int Prev = std::numeric_limits<int>::min();
  P[0] = -P[0];
  for (int I = 1; I < N; I++) {
    int Pos = P[I];
    int Neg = -P[I];
    if (Neg > P[I - 1]) {
      P[I] = Neg;
      continue;
    }
    if (Pos > P[I - 1]) continue;
    if (Neg > Prev) {
      Prev = P[I - 1];
      P[I] = Neg;
      continue;
    }
    if (Pos > Prev) {
      Prev = P[I - 1];
      continue;
    }
    cout << "NO\n";
    return;
  }
  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;
}
