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 namespace std;
const int INF = (int)1e9 + 7;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  for (int tc = 1; tc <= tt; tc++) {
    int sz;
    cin >> sz;
    vector<int> v(sz), sol;
    for (auto &x : v) {
      cin >> x;
    }
    int deja = 0;
    int initlast = INF;
    vector<int> guys = {INF, v[0]};
    int last = initlast;
    sol.push_back(v[0]);
    for (int i = 1; i < sz; i++) {
      if (guys.back() > guys[(int)guys.size() - 2]) {
        last = guys.back();
      }
      int x = v[i];
      if (x < guys.back()) {
        guys.back() = x;
        sol.push_back(x);
        continue;
      }
      if (1) {
        if (-x < guys.back()) {
          guys.back() = -x;
          sol.push_back(-x);
          continue;
        }
        if (x < last) {
          guys.push_back(x);
          sol.push_back(x);
          last = initlast;
          continue;
        }
      } else {
        if (x < last) {
          guys.push_back(x);
          sol.push_back(x);
          last = initlast;
          continue;
        }
        if (-x < guys.back()) {
          guys.back() = -x;
          sol.push_back(-x);
          continue;
        }
      }
      if (-x < last) {
        guys.push_back(-x);
        sol.push_back(-x);
        last = initlast;
        continue;
      }
      break;
    }
    if ((int)sol.size() != sz) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (auto &x : sol) {
      cout << -x << " ";
    }
    cout << "\n";
  }
  return 0;
}
