#include <bits/stdc++.h>
int _;
using namespace std;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 1e7 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
int n;
int a[N];
struct status {
  int pre;
  int x, y;
  int val;
} s[N];
int p = 0;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  s[p = 1] = {-1, -INF, -INF, -INF};
  vector<int> lst, cur;
  lst.push_back(1);
  for (int i = 1; i <= n; ++i) {
    for (int j : lst) {
      auto [pre, x, y, val] = s[j];
      for (int newx : {-a[i], a[i]})
        if (x < newx) {
          s[++p] = {j, newx, y, newx};
          cur.push_back(p);
        }
      for (int newy : {-a[i], a[i]})
        if (y < newy) {
          s[++p] = {j, x, newy, newy};
          cur.push_back(p);
        }
    }
    if (((int)(cur).size()) == 0) return (void)(cout << "NO" << endl);
    lst.clear();
    sort((cur).begin(), (cur).end(), [](int a, int b) {
      if (s[a].x != s[b].x) return s[a].x < s[b].x;
      return s[a].y < s[b].y;
    });
    for (int j : cur) {
      if ((((int)(lst).size()) == 0) || (s[j].y < s[lst.back()].y))
        lst.push_back(j);
    }
    cur.clear();
  }
  cout << "YES" << endl;
  vector<int> ans;
  int u = lst[0];
  while (s[u].pre != -1) {
    ans.push_back(s[u].val);
    u = s[u].pre;
  }
  reverse((ans).begin(), (ans).end());
  for (int i : ans) {
    cout << i << ' ';
  }
  cout << endl;
}
void main0() {
  int T;
  cin >> T;
  for (_ = 1; _ <= T; ++_) {
    solve();
  }
}

### 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>
int _;
using namespace std;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 1e7 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
int n;
int a[N];
struct status {
  int pre;
  int x, y;
  int val;
} s[N];
int p = 0;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  s[p = 1] = {-1, -INF, -INF, -INF};
  vector<int> lst, cur;
  lst.push_back(1);
  for (int i = 1; i <= n; ++i) {
    for (int j : lst) {
      auto [pre, x, y, val] = s[j];
      for (int newx : {-a[i], a[i]})
        if (x < newx) {
          s[++p] = {j, newx, y, newx};
          cur.push_back(p);
        }
      for (int newy : {-a[i], a[i]})
        if (y < newy) {
          s[++p] = {j, x, newy, newy};
          cur.push_back(p);
        }
    }
    if (((int)(cur).size()) == 0) return (void)(cout << "NO" << endl);
    lst.clear();
    sort((cur).begin(), (cur).end(), [](int a, int b) {
      if (s[a].x != s[b].x) return s[a].x < s[b].x;
      return s[a].y < s[b].y;
    });
    for (int j : cur) {
      if ((((int)(lst).size()) == 0) || (s[j].y < s[lst.back()].y))
        lst.push_back(j);
    }
    cur.clear();
  }
  cout << "YES" << endl;
  vector<int> ans;
  int u = lst[0];
  while (s[u].pre != -1) {
    ans.push_back(s[u].val);
    u = s[u].pre;
  }
  reverse((ans).begin(), (ans).end());
  for (int i : ans) {
    cout << i << ' ';
  }
  cout << endl;
}
void main0() {
  int T;
  cin >> T;
  for (_ = 1; _ <= T; ++_) {
    solve();
  }
}

```