#include <bits/stdc++.h>
using namespace std;
int p[1000005], dp[1000005][2][2], n;
struct Node {
  int i, j, k;
} last[1000005][2][2];
void Min(int &a, int b, Node c, Node &d) {
  if (a > b) a = b, d = c;
}
void Main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k) dp[i][j][k] = (int)1e9;
  dp[1][0][0] = -(int)1e9;
  dp[1][0][1] = -(int)1e9;
  for (int i = 1; i < n; ++i)
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k) {
        int a, b;
        if (j == 0) a = (k == 0 ? -p[i] : p[i]), b = dp[i][j][k];
        if (j == 1) a = dp[i][j][k], b = (k == 0 ? -p[i] : p[i]);
        if (p[i + 1] > a)
          Min(dp[i + 1][0][1], b, (Node){i, j, k}, last[i + 1][0][1]);
        else if (p[i + 1] > b)
          Min(dp[i + 1][1][1], a, (Node){i, j, k}, last[i + 1][1][1]);
        if (-p[i + 1] > a)
          Min(dp[i + 1][0][0], b, (Node){i, j, k}, last[i + 1][0][0]);
        else if (-p[i + 1] > b)
          Min(dp[i + 1][1][0], a, (Node){i, j, k}, last[i + 1][1][0]);
      }
  int nj = -1, nk = -1;
  for (int j = 0; j < 2; ++j)
    for (int k = 0; k < 2; ++k)
      if (dp[n][j][k] < (int)1e9) {
        nj = j;
        nk = k;
      }
  if (nj == -1) {
    puts("NO");
    return;
  }
  for (int i = n; i >= 1; --i) {
    if (!nk) p[i] = -p[i];
    int aj = last[i][nj][nk].j, ak = last[i][nj][nk].k;
    nj = aj;
    nk = ak;
  }
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d%c", p[i], (i == n ? '\n' : ' '));
}
int main() {
  int T;
  scanf("%d", &T);
  for (; T--;) Main();
}

### Prompt

Please formulate 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>
using namespace std;
int p[1000005], dp[1000005][2][2], n;
struct Node {
  int i, j, k;
} last[1000005][2][2];
void Min(int &a, int b, Node c, Node &d) {
  if (a > b) a = b, d = c;
}
void Main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k) dp[i][j][k] = (int)1e9;
  dp[1][0][0] = -(int)1e9;
  dp[1][0][1] = -(int)1e9;
  for (int i = 1; i < n; ++i)
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k) {
        int a, b;
        if (j == 0) a = (k == 0 ? -p[i] : p[i]), b = dp[i][j][k];
        if (j == 1) a = dp[i][j][k], b = (k == 0 ? -p[i] : p[i]);
        if (p[i + 1] > a)
          Min(dp[i + 1][0][1], b, (Node){i, j, k}, last[i + 1][0][1]);
        else if (p[i + 1] > b)
          Min(dp[i + 1][1][1], a, (Node){i, j, k}, last[i + 1][1][1]);
        if (-p[i + 1] > a)
          Min(dp[i + 1][0][0], b, (Node){i, j, k}, last[i + 1][0][0]);
        else if (-p[i + 1] > b)
          Min(dp[i + 1][1][0], a, (Node){i, j, k}, last[i + 1][1][0]);
      }
  int nj = -1, nk = -1;
  for (int j = 0; j < 2; ++j)
    for (int k = 0; k < 2; ++k)
      if (dp[n][j][k] < (int)1e9) {
        nj = j;
        nk = k;
      }
  if (nj == -1) {
    puts("NO");
    return;
  }
  for (int i = n; i >= 1; --i) {
    if (!nk) p[i] = -p[i];
    int aj = last[i][nj][nk].j, ak = last[i][nj][nk].k;
    nj = aj;
    nk = ak;
  }
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d%c", p[i], (i == n ? '\n' : ' '));
}
int main() {
  int T;
  scanf("%d", &T);
  for (; T--;) Main();
}

```