#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
long long t, n, a[MAX];
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  long long i, j, f1, f2, g1, g2, gg;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) scanf("%lld", a + i);
    if (n % 2 == 0) {
      for (i = 1; i <= n; i += 2) {
        f1 = f2 = 0;
        g1 = g2 = 1;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
        else
          printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
      }
      printf("\n");
    } else {
      for (i = 1; i <= n - 3; i += 2) {
        f1 = f2 = 0;
        g1 = g2 = 1;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
        else
          printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
      }
      if (a[i] == a[i + 1]) {
        f1 = f2 = 0;
        i++;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("-1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
        else
          printf("1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
        printf("\n");
        continue;
      }
      if (a[i] == -a[i + 1]) {
        f1 = f2 = 0;
        i++;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
        else
          printf("-1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
        printf("\n");
        continue;
      }
      long long kk = 1;
      if ((a[i + 1] < 0 && a[i] > 0) || (a[i + 1] > 0 && a[i] < 0)) {
        a[i + 1] += a[i];
        i++;
      } else {
        a[i + 1] -= a[i];
        i++;
        kk = -1;
      }
      f1 = f2 = 0;
      if (a[i] < 0) a[i] = -a[i], f1 = 1;
      if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
      gg = gcd(a[i], a[i + 1]);
      if (f1 ^ f2)
        printf("%lld %lld %lld ", a[i + 1] * kk / gg, a[i + 1] / gg, a[i] / gg);
      else
        printf("%lld %lld %lld ", -a[i + 1] * kk / gg, -a[i + 1] / gg,
               a[i] / gg);
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int MAX = 1e5 + 5;
long long t, n, a[MAX];
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  long long i, j, f1, f2, g1, g2, gg;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) scanf("%lld", a + i);
    if (n % 2 == 0) {
      for (i = 1; i <= n; i += 2) {
        f1 = f2 = 0;
        g1 = g2 = 1;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
        else
          printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
      }
      printf("\n");
    } else {
      for (i = 1; i <= n - 3; i += 2) {
        f1 = f2 = 0;
        g1 = g2 = 1;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
        else
          printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
      }
      if (a[i] == a[i + 1]) {
        f1 = f2 = 0;
        i++;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("-1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
        else
          printf("1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
        printf("\n");
        continue;
      }
      if (a[i] == -a[i + 1]) {
        f1 = f2 = 0;
        i++;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
        else
          printf("-1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
        printf("\n");
        continue;
      }
      long long kk = 1;
      if ((a[i + 1] < 0 && a[i] > 0) || (a[i + 1] > 0 && a[i] < 0)) {
        a[i + 1] += a[i];
        i++;
      } else {
        a[i + 1] -= a[i];
        i++;
        kk = -1;
      }
      f1 = f2 = 0;
      if (a[i] < 0) a[i] = -a[i], f1 = 1;
      if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
      gg = gcd(a[i], a[i + 1]);
      if (f1 ^ f2)
        printf("%lld %lld %lld ", a[i + 1] * kk / gg, a[i + 1] / gg, a[i] / gg);
      else
        printf("%lld %lld %lld ", -a[i + 1] * kk / gg, -a[i + 1] / gg,
               a[i] / gg);
      printf("\n");
    }
  }
  return 0;
}

```