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 N = (int)2e6 + 7;
const int INF = (int)1e9 + 7;
int n;
int a[N];
int b[N];
int aib[N];
void upd(int i, int x) {
  while (i < N) {
    aib[i] = max(aib[i], x);
    i += i & (-i);
  }
}
int get(int i) {
  int sol = 0;
  while (i) {
    sol = max(sol, aib[i]);
    i -= i & (-i);
  }
  return sol;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  for (int tc = 1; tc <= tt; tc++) {
    cin >> n;
    for (int i = 1; i <= 2 * n; i++) {
      aib[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[i] = -a[i];
      a[i] += n;
      b[i] += n + 1;
      assert(1 <= a[i] && a[i] <= 2 * n);
      assert(1 <= b[i] && b[i] <= 2 * n);
    }
    vector<int> sol;
    for (int i = 1; i <= n; i++) {
      if (get(a[i]) <= 1) {
        upd(a[i], get(a[i]) + 1);
        sol.push_back(a[i]);
        continue;
      }
      if (get(b[i]) <= 1) {
        upd(b[i], get(b[i]) + 1);
        sol.push_back(b[i]);
        continue;
      }
      break;
    }
    if ((int)sol.size() != n) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (auto &it : sol) {
      if (it <= n) {
        it -= (n + 1);
      } else {
        it -= n;
      }
      cout << it << " ";
    }
    cout << "\n";
  }
  return 0;
}
