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 + 1e3 + 7;
int T, n, a[N];
pair<int, int> c[N];
void add(int x, pair<int, int> v) {
  while (x <= n * 2 + 5) {
    c[x] = max(c[x], v);
    x += x & -x;
  }
}
pair<int, int> qry(int x) {
  pair<int, int> ret = {0, 0};
  while (x) {
    ret = max(ret, c[x]);
    x -= x & -x;
  }
  return ret;
}
int vis[N], last[N];
int main() {
  scanf("%d", &T);
  while (T--) {
    for (int i = 0; i <= n * 2 + 10; i++) c[i] = {0, 0};
    scanf("%d", &n);
    int ans = 0;
    int pos = -1;
    for (int i = 1; i <= n; i++) {
      vis[i] = 0;
      scanf("%d", &a[i]);
      pair<int, int> f0 = qry(a[i] - 1 + n + 1);
      f0.first++;
      pair<int, int> f1 = qry(-a[i] - 1 + n + 1);
      f1.first++;
      last[i * 2 - 1] = f0.second;
      last[i * 2] = f1.second;
      f0.second = i * 2 - 1;
      f1.second = i * 2;
      add(a[i] + n + 1, f0);
      add(-a[i] + n + 1, f1);
      if (f0.first > ans) ans = f0.first, pos = f0.second;
      if (f1.first > ans) ans = f1.first, pos = f1.second;
    }
    for (int i = pos; i; i = last[i]) vis[(i + 1) / 2] = i % 2 ? 1 : -1;
    int lst = -n + 1;
    int ok = 1;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) {
        if (-a[i] > lst) a[i] = -a[i];
        if (a[i] < lst) ok = 0;
        lst = a[i];
      }
    for (int i = 1; i <= n; i++)
      if (vis[i]) a[i] *= vis[i];
    puts(ok ? "YES" : "NO");
    if (ok) {
      for (int i = 1; i <= n; i++) printf("%d%c", a[i], " \n"[i == n]);
    }
  }
}
