#include <bits/stdc++.h>
using namespace std;
long long dp[1000006][2];
long long pre[1000006][2];
void chkMinLog(long long& goal, long long t1, bool t2, bool curr) {
  if (dp[t1][t2] > goal) {
    dp[t1][t2] = goal;
    pre[t1][t2] = curr;
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n + 2);
  for (long long i = (1); i <= (n); i++) cin >> a[i];
  for (long long i = (1); i <= (n); i++)
    for (long long j = (0); j <= (1); j++) dp[i][j] = 1e18;
  a[0] = n + 1;
  dp[0][0] = dp[0][1] = -n - 1;
  for (long long i = (0); i <= (n - 1); i++)
    for (long long isPos = (0); isPos <= (1); isPos++) {
      long long sv = (isPos ? 1 : -1) * a[i], v = dp[i][isPos];
      if (-a[i + 1] > v) chkMinLog(sv, i + 1, false, isPos);
      if (a[i + 1] > v) chkMinLog(sv, i + 1, true, isPos);
      if (-a[i + 1] > sv) chkMinLog(v, i + 1, false, isPos);
      if (a[i + 1] > sv) chkMinLog(v, i + 1, true, isPos);
    }
  if (dp[n][0] <= n || dp[n][1] <= n) {
    cout << "YES" << endl;
    long long ti = n, ts = 0;
    if (dp[n][1] <= n) ts = 1;
    while (ti >= 1) {
      if (ts == 0) a[ti] = -a[ti];
      ts = pre[ti][ts], ti--;
    }
    for (long long i = (1); i <= (n); i++) cout << a[i] << " ";
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long dp[1000006][2];
long long pre[1000006][2];
void chkMinLog(long long& goal, long long t1, bool t2, bool curr) {
  if (dp[t1][t2] > goal) {
    dp[t1][t2] = goal;
    pre[t1][t2] = curr;
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n + 2);
  for (long long i = (1); i <= (n); i++) cin >> a[i];
  for (long long i = (1); i <= (n); i++)
    for (long long j = (0); j <= (1); j++) dp[i][j] = 1e18;
  a[0] = n + 1;
  dp[0][0] = dp[0][1] = -n - 1;
  for (long long i = (0); i <= (n - 1); i++)
    for (long long isPos = (0); isPos <= (1); isPos++) {
      long long sv = (isPos ? 1 : -1) * a[i], v = dp[i][isPos];
      if (-a[i + 1] > v) chkMinLog(sv, i + 1, false, isPos);
      if (a[i + 1] > v) chkMinLog(sv, i + 1, true, isPos);
      if (-a[i + 1] > sv) chkMinLog(v, i + 1, false, isPos);
      if (a[i + 1] > sv) chkMinLog(v, i + 1, true, isPos);
    }
  if (dp[n][0] <= n || dp[n][1] <= n) {
    cout << "YES" << endl;
    long long ti = n, ts = 0;
    if (dp[n][1] <= n) ts = 1;
    while (ti >= 1) {
      if (ts == 0) a[ti] = -a[ti];
      ts = pre[ti][ts], ti--;
    }
    for (long long i = (1); i <= (n); i++) cout << a[i] << " ";
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}

```