#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
bool uin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
template <class T>
bool uax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int inf = 1e9;
void solve() {
  int n;
  cin >> n;
  vector<int> p(n);
  for (auto& x : p) cin >> x;
  p.insert(begin(p), -inf);
  vector<array<bool, 2>> par(n + 1);
  array<int, 2> dp;
  fill(begin(dp), end(dp), -inf);
  for (int i = 1; i <= n; ++i) {
    array<int, 2> nxt;
    fill(begin(nxt), end(nxt), inf);
    for (int x = -p[i], j = 0; j < 2; ++j, x = -x) {
      if (x > +p[i - 1] && uin(nxt[j], dp[1])) par[i][j] = 1;
      if (x > dp[1] && uin(nxt[j], +p[i - 1])) par[i][j] = 1;
      if (x > -p[i - 1] && uin(nxt[j], dp[0])) par[i][j] = 0;
      if (x > dp[0] && uin(nxt[j], -p[i - 1])) par[i][j] = 0;
    }
    dp = move(nxt);
  }
  if (min(dp[0], dp[1]) > n) {
    cout << "NO\n";
    return;
  }
  vector<int> res(n);
  for (int i = n, state = dp[1] <= n; i > 0; --i) {
    res[i - 1] = state ? +p[i] : -p[i];
    state = par[i][state];
  }
  cout << "YES\n";
  for (auto x : res) cout << x << ' ';
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
using ll = long long;
using ld = long double;
template <class T>
bool uin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
template <class T>
bool uax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int inf = 1e9;
void solve() {
  int n;
  cin >> n;
  vector<int> p(n);
  for (auto& x : p) cin >> x;
  p.insert(begin(p), -inf);
  vector<array<bool, 2>> par(n + 1);
  array<int, 2> dp;
  fill(begin(dp), end(dp), -inf);
  for (int i = 1; i <= n; ++i) {
    array<int, 2> nxt;
    fill(begin(nxt), end(nxt), inf);
    for (int x = -p[i], j = 0; j < 2; ++j, x = -x) {
      if (x > +p[i - 1] && uin(nxt[j], dp[1])) par[i][j] = 1;
      if (x > dp[1] && uin(nxt[j], +p[i - 1])) par[i][j] = 1;
      if (x > -p[i - 1] && uin(nxt[j], dp[0])) par[i][j] = 0;
      if (x > dp[0] && uin(nxt[j], -p[i - 1])) par[i][j] = 0;
    }
    dp = move(nxt);
  }
  if (min(dp[0], dp[1]) > n) {
    cout << "NO\n";
    return;
  }
  vector<int> res(n);
  for (int i = n, state = dp[1] <= n; i > 0; --i) {
    res[i - 1] = state ? +p[i] : -p[i];
    state = par[i][state];
  }
  cout << "YES\n";
  for (auto x : res) cout << x << ' ';
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}

```