#include <bits/stdc++.h>
const int INF = 2e9;
inline int Max(const int &a, const int &b) { return a > b ? a : b; }
inline int Min(const int &a, const int &b) { return a > b ? b : a; }
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
using namespace std;
const int N = 1e6 + 7;
int T, n;
int p[N], b[N];
struct jgt {
  int j, k, ls, tp;
} q[N * 4];
int cnt;
vector<int> f[N];
void dfs(int id, int i) {
  if (!i) return;
  b[i] = q[id].tp * p[i];
  dfs(q[id].ls, i - 1);
}
int a, j, k;
int ax, _ax, xa, x_a;
int axid, _axid, xaid, x_aid;
void solve() {
  n = read();
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= n; ++i) f[i].clear();
  q[cnt = 1] = {-p[1], -INF, 0, -1};
  f[1].push_back(1);
  for (int i = 2; i <= n; ++i) {
    if (f[i - 1].empty()) {
      puts("NO");
      return;
    }
    a = p[i];
    ax = _ax = xa = x_a = INF;
    axid = _axid = xaid = x_aid = 0;
    for (int id : f[i - 1]) {
      j = q[id].j, k = q[id].k;
      if (k < -a) {
        if (j < -a) {
          if (k < _ax) _ax = k, _axid = id;
        } else {
          if (j < x_a) x_a = j, x_aid = id;
        }
      }
      if (k < a) {
        if (j < a) {
          if (k < ax) ax = k, axid = id;
        } else {
          if (j < xa) xa = j, xaid = id;
        }
      }
    }
    if (ax < INF) {
      q[++cnt] = {a, ax, axid, 1};
      f[i].push_back(cnt);
    }
    if (xa < INF) {
      q[++cnt] = {xa, a, xaid, 1};
      f[i].push_back(cnt);
    }
    if (_ax < INF) {
      q[++cnt] = {-a, _ax, _axid, -1};
      f[i].push_back(cnt);
    }
    if (x_a < INF) {
      q[++cnt] = {x_a, -a, x_aid, -1};
      f[i].push_back(cnt);
    }
  }
  if (f[n].empty()) {
    puts("NO");
  } else {
    puts("YES");
    dfs(f[n][0], n);
    for (int i = 1; i <= n; ++i) printf("%d ", b[i]);
    puts("");
  }
}
int main() {
  T = read();
  for (int i = 1; i <= T; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
const int INF = 2e9;
inline int Max(const int &a, const int &b) { return a > b ? a : b; }
inline int Min(const int &a, const int &b) { return a > b ? b : a; }
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
using namespace std;
const int N = 1e6 + 7;
int T, n;
int p[N], b[N];
struct jgt {
  int j, k, ls, tp;
} q[N * 4];
int cnt;
vector<int> f[N];
void dfs(int id, int i) {
  if (!i) return;
  b[i] = q[id].tp * p[i];
  dfs(q[id].ls, i - 1);
}
int a, j, k;
int ax, _ax, xa, x_a;
int axid, _axid, xaid, x_aid;
void solve() {
  n = read();
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= n; ++i) f[i].clear();
  q[cnt = 1] = {-p[1], -INF, 0, -1};
  f[1].push_back(1);
  for (int i = 2; i <= n; ++i) {
    if (f[i - 1].empty()) {
      puts("NO");
      return;
    }
    a = p[i];
    ax = _ax = xa = x_a = INF;
    axid = _axid = xaid = x_aid = 0;
    for (int id : f[i - 1]) {
      j = q[id].j, k = q[id].k;
      if (k < -a) {
        if (j < -a) {
          if (k < _ax) _ax = k, _axid = id;
        } else {
          if (j < x_a) x_a = j, x_aid = id;
        }
      }
      if (k < a) {
        if (j < a) {
          if (k < ax) ax = k, axid = id;
        } else {
          if (j < xa) xa = j, xaid = id;
        }
      }
    }
    if (ax < INF) {
      q[++cnt] = {a, ax, axid, 1};
      f[i].push_back(cnt);
    }
    if (xa < INF) {
      q[++cnt] = {xa, a, xaid, 1};
      f[i].push_back(cnt);
    }
    if (_ax < INF) {
      q[++cnt] = {-a, _ax, _axid, -1};
      f[i].push_back(cnt);
    }
    if (x_a < INF) {
      q[++cnt] = {x_a, -a, x_aid, -1};
      f[i].push_back(cnt);
    }
  }
  if (f[n].empty()) {
    puts("NO");
  } else {
    puts("YES");
    dfs(f[n][0], n);
    for (int i = 1; i <= n; ++i) printf("%d ", b[i]);
    puts("");
  }
}
int main() {
  T = read();
  for (int i = 1; i <= T; ++i) {
    solve();
  }
  return 0;
}

```