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 = 1e6 + 5, inf = 1e9;
int n, a[N], f[N][2], fro[N][2], ans[N];
void upd(int i, int j, int k, int x) {
  if (f[i][j] == inf) f[i][j] = x, fro[i][j] = k;
  if (f[i][j] > x) f[i][j] = x, fro[i][j] = k;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T, j, x, y, bf;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = (1); i <= (n); ++i) cin >> a[i];
    f[1][0] = f[1][1] = -inf;
    for (int i = (2); i <= (n); ++i) {
      f[i][0] = f[i][1] = inf;
      j = 1;
      x = -a[i];
      for (int k = (0); k <= (1); ++k) {
        if (f[i - 1][k] == inf) continue;
        y = (k ? -a[i - 1] : a[i - 1]);
        bf = (f[i - 1][k] != y ? y : (fro[i - 1][k] ? -a[i - 2] : a[i - 2]));
        if (x > f[i - 1][k]) {
          if (x > bf)
            upd(i, j, k, f[i - 1][k]);
          else
            upd(i, j, k, x);
        }
      }
      j = 0;
      x = a[i];
      for (int k = (0); k <= (1); ++k) {
        if (f[i - 1][k] == inf) continue;
        y = (k ? -a[i - 1] : a[i - 1]);
        bf = (f[i - 1][k] != y ? y : (fro[i - 1][k] ? -a[i - 2] : a[i - 2]));
        if (x > f[i - 1][k]) {
          if (x > bf)
            upd(i, j, k, f[i - 1][k]);
          else
            upd(i, j, k, x);
        }
      }
    }
    if (f[n][0] != inf)
      j = 0;
    else if (f[n][1] != inf)
      j = 1;
    else
      j = -1;
    if (j == -1) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    for (int i = (n); i >= (1); --i) {
      ans[i] = (j ? -a[i] : a[i]);
      j = fro[i][j];
    }
    for (int i = (1); i <= (n); ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0;
}
