#include <bits/stdc++.h>
using namespace std;
void solve(long long nn) {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> vp;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[i] = x;
    vp.push_back({x, i});
  }
  sort(vp.begin(), vp.end());
  long long second = 0;
  long long iid = -1;
  long long kk = INT_MAX;
  for (long long i = 0; i < n; i++) {
    long long val = vp[i].first;
    if (abs(val) < kk) {
      kk = abs(val);
      iid = i;
    }
  }
  long long k = vp[iid].first;
  long long ans[n];
  for (long long i = n - 1; i >= 0; i--) {
    if (i == iid) continue;
    pair<long long, long long> p = vp[i];
    long long id = p.second;
    long long val = p.first;
    long long zz = abs(val);
    if (i == 1) {
      if (iid == 0) {
        if (second > 0) {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        }
      } else {
        if (second > 0) {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        }
      }
    } else if (i == 0 and iid) {
      if (second > 0) {
        if (val > 0) {
          ans[id] = kk;
          second += zz * kk;
        } else {
          ans[id] = -kk;
          second += zz * kk;
        }
      } else {
        if (val > 0) {
          ans[id] = -1 * kk;
          second -= zz * kk;
        } else {
          ans[id] = kk;
          second -= zz * kk;
        }
      }
    } else {
      if (second > 0) {
        if (val > 0) {
          ans[id] = -1 * kk;
          second -= zz * kk;
        } else {
          ans[id] = kk;
          second -= zz * kk;
        }
      } else {
        if (val > 0) {
          ans[id] = kk;
          second += zz * kk;
        } else {
          ans[id] = -kk;
          second += zz * kk;
        }
      }
    }
  }
  ans[vp[iid].second] = -1 * second / vp[iid].first;
  long long ss = 0;
  for (long long i = 0; i < n; i++) {
    ss += ans[i] * a[i];
    cout << ans[i] << " ";
  }
  cout << "\n";
  return;
}
int main() {
  std::cout << std::fixed;
  std::cout << std::setprecision(15);
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Vupsen and Pupsen were gifted an integer array. Since Vupsen doesn't like the number 0, he threw away all numbers equal to 0 from the array. As a result, he got an array a of length n.

Pupsen, on the contrary, likes the number 0 and he got upset when he saw the array without zeroes. To cheer Pupsen up, Vupsen decided to come up with another array b of length n such that ∑_{i=1}^{n}a_i ⋅ b_i=0. Since Vupsen doesn't like number 0, the array b must not contain numbers equal to 0. Also, the numbers in that array must not be huge, so the sum of their absolute values cannot exceed 10^9. Please help Vupsen to find any such array b!

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2 ⋅ t lines contain the description of test cases. The description of each test case consists of two lines.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the length of the array.

The second line contains n integers a_1, a_2, …, a_n (-10^4 ≤ a_i ≤ 10^4, a_i ≠ 0) — the elements of the array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print n integers b_1, b_2, …, b_n — elements of the array b (|b_1|+|b_2|+… +|b_n| ≤ 10^9, b_i ≠ 0, ∑_{i=1}^{n}a_i ⋅ b_i=0).

It can be shown that the answer always exists.

Example

Input


3
2
5 5
5
5 -2 10 -9 4
7
1 2 3 4 5 6 7


Output


1 -1
-1 5 1 -1 -1
-10 2 2 -3 5 -1 -1

Note

In the first test case, 5 ⋅ 1 + 5 ⋅ (-1)=5-5=0. You could also print 3 -3, for example, since 5 ⋅ 3 + 5 ⋅ (-3)=15-15=0

In the second test case, 5 ⋅ (-1) + (-2) ⋅ 5 + 10 ⋅ 1 + (-9) ⋅ (-1) + 4 ⋅ (-1)=-5-10+10+9-4=0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve(long long nn) {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> vp;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[i] = x;
    vp.push_back({x, i});
  }
  sort(vp.begin(), vp.end());
  long long second = 0;
  long long iid = -1;
  long long kk = INT_MAX;
  for (long long i = 0; i < n; i++) {
    long long val = vp[i].first;
    if (abs(val) < kk) {
      kk = abs(val);
      iid = i;
    }
  }
  long long k = vp[iid].first;
  long long ans[n];
  for (long long i = n - 1; i >= 0; i--) {
    if (i == iid) continue;
    pair<long long, long long> p = vp[i];
    long long id = p.second;
    long long val = p.first;
    long long zz = abs(val);
    if (i == 1) {
      if (iid == 0) {
        if (second > 0) {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        }
      } else {
        if (second > 0) {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        }
      }
    } else if (i == 0 and iid) {
      if (second > 0) {
        if (val > 0) {
          ans[id] = kk;
          second += zz * kk;
        } else {
          ans[id] = -kk;
          second += zz * kk;
        }
      } else {
        if (val > 0) {
          ans[id] = -1 * kk;
          second -= zz * kk;
        } else {
          ans[id] = kk;
          second -= zz * kk;
        }
      }
    } else {
      if (second > 0) {
        if (val > 0) {
          ans[id] = -1 * kk;
          second -= zz * kk;
        } else {
          ans[id] = kk;
          second -= zz * kk;
        }
      } else {
        if (val > 0) {
          ans[id] = kk;
          second += zz * kk;
        } else {
          ans[id] = -kk;
          second += zz * kk;
        }
      }
    }
  }
  ans[vp[iid].second] = -1 * second / vp[iid].first;
  long long ss = 0;
  for (long long i = 0; i < n; i++) {
    ss += ans[i] * a[i];
    cout << ans[i] << " ";
  }
  cout << "\n";
  return;
}
int main() {
  std::cout << std::fixed;
  std::cout << std::setprecision(15);
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

```