#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using ull = unsigned long long;
using db = long double;
const int N = 1e6 + 5, LG = 18, MOD = 998244353;
const long double PI = acos(-1);
int dp[N][2];
int par[N][2];
void doWork() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < 2; j++) dp[i][j] = INT_MAX;
  }
  dp[0][0] = dp[0][1] = INT_MIN;
  auto upd = [&](int i, int j, int k, int p) {
    if (dp[i][j] > k) {
      dp[i][j] = k;
      par[i][j] = p;
    }
  };
  for (int i = 0; i < n - 1; i++)
    for (int x = 0; x < 2; x++)
      for (int y = 0; y < 2; y++) {
        int val1 = (x ? -1 : 1) * a[i];
        int val2 = (y ? -1 : 1) * a[i + 1];
        if (val1 <= val2) {
          upd(i + 1, y, dp[i][x], x);
        }
        if (dp[i][x] <= val2) {
          upd(i + 1, y, val1, x);
        }
      }
  if (dp[n - 1][0] == INT_MAX) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    int cur = 0;
    for (int i = n - 1; i >= 0; --i) {
      if (cur) a[i] *= -1;
      cur = par[i][cur];
    }
    for (int i = 0; i < n; i++) cout << a[i] << " \n"[i + 1 == n];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    doWork();
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using ull = unsigned long long;
using db = long double;
const int N = 1e6 + 5, LG = 18, MOD = 998244353;
const long double PI = acos(-1);
int dp[N][2];
int par[N][2];
void doWork() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < 2; j++) dp[i][j] = INT_MAX;
  }
  dp[0][0] = dp[0][1] = INT_MIN;
  auto upd = [&](int i, int j, int k, int p) {
    if (dp[i][j] > k) {
      dp[i][j] = k;
      par[i][j] = p;
    }
  };
  for (int i = 0; i < n - 1; i++)
    for (int x = 0; x < 2; x++)
      for (int y = 0; y < 2; y++) {
        int val1 = (x ? -1 : 1) * a[i];
        int val2 = (y ? -1 : 1) * a[i + 1];
        if (val1 <= val2) {
          upd(i + 1, y, dp[i][x], x);
        }
        if (dp[i][x] <= val2) {
          upd(i + 1, y, val1, x);
        }
      }
  if (dp[n - 1][0] == INT_MAX) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    int cur = 0;
    for (int i = n - 1; i >= 0; --i) {
      if (cur) a[i] *= -1;
      cur = par[i][cur];
    }
    for (int i = 0; i < n; i++) cout << a[i] << " \n"[i + 1 == n];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    doWork();
  }
  return 0;
}

```