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 Maxn = 1000005;
const int Inf = 1000000000;
int T;
int n;
int p[Maxn];
bool tk[Maxn];
int res[Maxn];
void Print() {
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d%c", res[i], i + 1 <= n ? ' ' : '\n');
}
bool Positive() {
  fill(tk, tk + n + 1, false);
  int pnt = n;
  int mn = Inf;
  for (int i = n; i > 0; i--) {
    while (tk[pnt]) pnt--;
    if (p[i] == pnt)
      res[i] = p[i];
    else if (p[i] < mn)
      res[i] = p[i];
    else if (-p[i] < mn)
      res[i] = -p[i];
    else
      return false;
    tk[p[i]] = true;
    mn = min(mn, res[i]);
  }
  return true;
}
bool Negative() {
  fill(tk, tk + n + 1, false);
  int pnt = n;
  int mx = -Inf;
  for (int i = 1; i <= n; i++) {
    while (tk[pnt]) pnt--;
    if (p[i] == pnt)
      res[i] = -p[i];
    else if (-p[i] > mx)
      res[i] = -p[i];
    else if (p[i] > mx)
      res[i] = p[i];
    else
      return false;
    tk[p[i]] = true;
    mx = max(mx, res[i]);
  }
  return true;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
    if (Positive())
      Print();
    else if (Negative())
      Print();
    else
      printf("NO\n");
  }
  return 0;
}
