#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int t, n;
int a[N];
int dp[N][2][2];
pair<int, pair<int, int> > be[N][2][2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i)
      cin >> a[i], dp[i][0][0] = dp[i][1][0] = dp[i][1][1] = dp[i][0][1] = 1e9,
                   be[i][0][0] = be[i][1][0] = be[i][1][1] = be[i][0][1] =
                       make_pair(0, make_pair(0, 0));
    be[1][0][1] = be[1][1][1] = make_pair(-1, make_pair(0, 0));
    dp[1][0][1] = -1e9;
    dp[1][1][1] = -1e9;
    for (int i = 2; i <= n; ++i)
      for (int j = 0; j <= 1; ++j)
        for (int op1 = 0; op1 <= 1; ++op1)
          for (int op2 = 0; op2 <= 1; ++op2) {
            int now = j == 0 ? -a[i] : a[i];
            int from = op1 == 0 ? -a[i - 1] : a[i - 1];
            if (be[i - 1][op1][op2].first == 0) continue;
            if (op2 == 0) {
              if (now < from) continue;
              if (now > dp[i - 1][op1][op2]) {
                if (from < dp[i][j][1])
                  dp[i][j][1] = from,
                  be[i][j][1] = make_pair(i - 1, make_pair(op1, op2));
              } else {
                if (dp[i - 1][op1][op2] < dp[i][j][0])
                  dp[i][j][0] = dp[i - 1][op1][op2],
                  be[i][j][0] = make_pair(i - 1, make_pair(op1, op2));
              }
            } else {
              if (now < dp[i - 1][op1][op2]) continue;
              if (now > from) {
                if (dp[i - 1][op1][op2] < dp[i][j][1])
                  dp[i][j][1] = dp[i - 1][op1][op2],
                  be[i][j][1] = make_pair(i - 1, make_pair(op1, op2));
              } else {
                if (from < dp[i][j][0])
                  dp[i][j][0] = from,
                  be[i][j][0] = make_pair(i - 1, make_pair(op1, op2));
              }
            }
          }
    int op1, op2, flag = 0;
    for (op1 = 0; op1 <= 1 && flag == 0; ++op1) {
      for (op2 = 0; op2 <= 1 && flag == 0; ++op2)
        if (be[n][op1][op2].first != 0) {
          flag = 1;
          break;
        }
      if (flag) break;
    }
    if (op1 > 1)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = n; i; --i) {
        pair<int, pair<int, int> > from = be[i][op1][op2];
        if (op1 == 0) a[i] = -a[i];
        op1 = from.second.first;
        op2 = from.second.second;
      }
      for (int i = 1; i <= n; ++i) cout << a[i] << " ";
      cout << "\n";
    }
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 1e6 + 10;
int t, n;
int a[N];
int dp[N][2][2];
pair<int, pair<int, int> > be[N][2][2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i)
      cin >> a[i], dp[i][0][0] = dp[i][1][0] = dp[i][1][1] = dp[i][0][1] = 1e9,
                   be[i][0][0] = be[i][1][0] = be[i][1][1] = be[i][0][1] =
                       make_pair(0, make_pair(0, 0));
    be[1][0][1] = be[1][1][1] = make_pair(-1, make_pair(0, 0));
    dp[1][0][1] = -1e9;
    dp[1][1][1] = -1e9;
    for (int i = 2; i <= n; ++i)
      for (int j = 0; j <= 1; ++j)
        for (int op1 = 0; op1 <= 1; ++op1)
          for (int op2 = 0; op2 <= 1; ++op2) {
            int now = j == 0 ? -a[i] : a[i];
            int from = op1 == 0 ? -a[i - 1] : a[i - 1];
            if (be[i - 1][op1][op2].first == 0) continue;
            if (op2 == 0) {
              if (now < from) continue;
              if (now > dp[i - 1][op1][op2]) {
                if (from < dp[i][j][1])
                  dp[i][j][1] = from,
                  be[i][j][1] = make_pair(i - 1, make_pair(op1, op2));
              } else {
                if (dp[i - 1][op1][op2] < dp[i][j][0])
                  dp[i][j][0] = dp[i - 1][op1][op2],
                  be[i][j][0] = make_pair(i - 1, make_pair(op1, op2));
              }
            } else {
              if (now < dp[i - 1][op1][op2]) continue;
              if (now > from) {
                if (dp[i - 1][op1][op2] < dp[i][j][1])
                  dp[i][j][1] = dp[i - 1][op1][op2],
                  be[i][j][1] = make_pair(i - 1, make_pair(op1, op2));
              } else {
                if (from < dp[i][j][0])
                  dp[i][j][0] = from,
                  be[i][j][0] = make_pair(i - 1, make_pair(op1, op2));
              }
            }
          }
    int op1, op2, flag = 0;
    for (op1 = 0; op1 <= 1 && flag == 0; ++op1) {
      for (op2 = 0; op2 <= 1 && flag == 0; ++op2)
        if (be[n][op1][op2].first != 0) {
          flag = 1;
          break;
        }
      if (flag) break;
    }
    if (op1 > 1)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = n; i; --i) {
        pair<int, pair<int, int> > from = be[i][op1][op2];
        if (op1 == 0) a[i] = -a[i];
        op1 = from.second.first;
        op2 = from.second.second;
      }
      for (int i = 1; i <= n; ++i) cout << a[i] << " ";
      cout << "\n";
    }
  }
}

```