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 = 1e9;
const int MOD = 1e9 + 7;
const int N = 1e6 + 100;
void tmax(int& a, int b) { a = max(a, b); }
void tmin(int& a, int b) { a = min(a, b); }
int a[N], pos[N], ans[N];
bool done[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  fill(done, done + n, 0);
  int ml, cl, md, cd, tl, dr;
  ml = cl = -INF;
  md = cd = +INF;
  tl = 0;
  dr = n - 1;
  for (int i = n; i >= 1 && tl <= dr; i--) {
    if (done[i]) continue;
    int bm = pos[i];
    bool lef = 1;
    int apl = ml;
    for (int j = tl; j < bm; j++) {
      if (a[j] < apl) {
        lef = 0;
        break;
      }
      if (-a[j] > apl) {
        ans[j] = -a[j];
        tmax(apl, ans[j]);
      } else {
        ans[j] = a[j];
        tmax(apl, ans[j]);
      }
    }
    bool rig = 1;
    int apr = cd;
    for (int j = dr; j > bm; j--) {
      if (-a[j] > apr) {
        rig = 0;
        break;
      }
      if (a[j] < apr) {
        ans[j] = a[j];
        tmin(apr, ans[j]);
      } else {
        ans[j] = -a[j];
        tmin(apr, ans[j]);
      }
    }
    if (-i > apr) lef = 0;
    if (i < apl) rig = 0;
    if (!lef && !rig) {
      cout << "NO" << '\n';
      return;
    }
    if (lef) {
      ans[bm] = -i;
      for (int j = tl; j <= bm; j++) done[a[j]] = 1;
      cl = -i;
      ml = apl;
      tl = bm + 1;
    } else {
      ans[bm] = i;
      for (int j = dr; j >= bm; j--) done[a[j]] = 1;
      md = i;
      cd = apr;
      dr = bm - 1;
    }
  }
  cout << "YES" << '\n';
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}
