#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
using ll = long long;
template <typename F>
void multitest(F func) {
  int t;
  cin >> t;
  while (t--) func();
}
void report(int ok) { cout << (ok ? "YES" : "NO") << '\n'; }
const int N = 1e6 + 16;
int dp[N][2];
int p[N][2];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  multitest([&]() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &i : a) cin >> i;
    for (int i = 0; i <= n; i++) dp[i][0] = dp[i][1] = 1 << 30;
    dp[0][0] = dp[0][1] = -(1 << 30);
    auto upd = [&](int i, int j, int k, int U) {
      if (dp[i][j] > k) {
        dp[i][j] = k;
        p[i][j] = U;
      }
    };
    for (int i = 0; i + 1 < n; i++) {
      for (int z = 0; z < 2; z++) {
        int v = z ? -a[i] : a[i];
        for (int f = 0; f < 2; f++) {
          int vv = f ? -a[i + 1] : a[i + 1];
          if (v <= vv) {
            upd(i + 1, f, dp[i][z], z);
          }
          if (dp[i][z] <= vv) {
            upd(i + 1, f, v, z);
          }
        }
      }
    }
    if (dp[n - 1][0] == (1 << 30)) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      int t = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (t) a[i] *= -1;
        t = p[i][t];
      }
      for (auto i : a) cout << i << " ";
      cout << '\n';
    }
  });
}

### Prompt

Your task is to create a Cpp solution to 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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
using ll = long long;
template <typename F>
void multitest(F func) {
  int t;
  cin >> t;
  while (t--) func();
}
void report(int ok) { cout << (ok ? "YES" : "NO") << '\n'; }
const int N = 1e6 + 16;
int dp[N][2];
int p[N][2];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  multitest([&]() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &i : a) cin >> i;
    for (int i = 0; i <= n; i++) dp[i][0] = dp[i][1] = 1 << 30;
    dp[0][0] = dp[0][1] = -(1 << 30);
    auto upd = [&](int i, int j, int k, int U) {
      if (dp[i][j] > k) {
        dp[i][j] = k;
        p[i][j] = U;
      }
    };
    for (int i = 0; i + 1 < n; i++) {
      for (int z = 0; z < 2; z++) {
        int v = z ? -a[i] : a[i];
        for (int f = 0; f < 2; f++) {
          int vv = f ? -a[i + 1] : a[i + 1];
          if (v <= vv) {
            upd(i + 1, f, dp[i][z], z);
          }
          if (dp[i][z] <= vv) {
            upd(i + 1, f, v, z);
          }
        }
      }
    }
    if (dp[n - 1][0] == (1 << 30)) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      int t = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (t) a[i] *= -1;
        t = p[i][t];
      }
      for (auto i : a) cout << i << " ";
      cout << '\n';
    }
  });
}

```