#include <bits/stdc++.h>
using namespace std;
void print() { cout << '\n'; }
template <typename T, typename... Types>
void print(T var1, Types... var2) {
  cout << var1 << ' ';
  print(var2...);
}
template <typename T>
void print(vector<T> &var1) {
  for (auto &i : var1) cout << i << ' ';
  cout << '\n';
}
template <typename T>
void print(T start, T end) {
  for (; start != end; start++) cout << *start << ' ';
  cout << '\n';
}
template <typename T>
void print(pair<T, T> &var1) {
  cout << var1.first << ' ' << var1.second << ' ';
  cout << '\n';
}
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename... Types>
void read(T &var1, Types &...var2) {
  cin >> var1;
  read(var2...);
}
template <typename T>
void read(vector<T> &var1) {
  for (auto &i : var1) read(i);
}
template <typename T>
void read(T start, T end) {
  for (; start != end; start++) cin >> *start;
}
template <typename T>
void read(pair<T, T> &var1) {
  cin >> var1.first >> var1.second;
}
template <typename T>
T Max(T x) {
  return x;
}
template <typename T, typename... Types>
T Max(T var1, Types... var2) {
  return var1 >= Max(var2...) ? var1 : Max(var2...);
}
template <typename T>
T Min(T x) {
  return x;
}
template <typename T, typename... Types>
T Min(T var1, Types... var2) {
  return var1 <= Min(var2...) ? var1 : Min(var2...);
}
long long INF = 1000000000000000005LL;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}
void solve() {
  long long l, r, m, a, b, c, n;
  read(l, r, m);
  long long val = r - l;
  for (long long i = l; i < r + 1; i++) {
    long long rem = m % i;
    if (rem <= val || (i - rem) <= val) {
      long long xy = (m - rem) / i;
      long long yz = (m + i - rem) / i;
      if ((xy > 0 && rem <= val)) {
        a = i;
        c = l;
        b = c + rem;
        print(a, b, c);
        return;
      }
      if (yz > 0 && (i - rem) <= val) {
        a = i;
        b = l;
        c = b + i - rem;
        print(a, b, c);
        return;
      }
    }
  }
}

### Prompt

Create a solution in Cpp for the following problem:

Pasha loves to send strictly positive integers to his friends. Pasha cares about security, therefore when he wants to send an integer n, he encrypts it in the following way: he picks three integers a, b and c such that l ≤ a,b,c ≤ r, and then he computes the encrypted value m = n ⋅ a + b - c.

Unfortunately, an adversary intercepted the values l, r and m. Is it possible to recover the original values of a, b and c from this information? More formally, you are asked to find any values of a, b and c such that

  * a, b and c are integers, 
  * l ≤ a, b, c ≤ r, 
  * there exists a strictly positive integer n, such that n ⋅ a + b - c = m. 

Input

The first line contains the only integer t (1 ≤ t ≤ 20) — the number of test cases. The following t lines describe one test case each.

Each test case consists of three integers l, r and m (1 ≤ l ≤ r ≤ 500 000, 1 ≤ m ≤ 10^{10}). The numbers are such that the answer to the problem exists.

Output

For each test case output three integers a, b and c such that, l ≤ a, b, c ≤ r and there exists a strictly positive integer n such that n ⋅ a + b - c = m. It is guaranteed that there is at least one possible solution, and you can output any possible combination if there are multiple solutions.

Example

Input


2
4 6 13
2 3 1


Output


4 6 5
2 2 3

Note

In the first example n = 3 is possible, then n ⋅ 4 + 6 - 5 = 13 = m. Other possible solutions include: a = 4, b = 5, c = 4 (when n = 3); a = 5, b = 4, c = 6 (when n = 3); a = 6, b = 6, c = 5 (when n = 2); a = 6, b = 5, c = 4 (when n = 2).

In the second example the only possible case is n = 1: in this case n ⋅ 2 + 2 - 3 = 1 = m. Note that, n = 0 is not possible, since in that case n is not a strictly positive integer.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void print() { cout << '\n'; }
template <typename T, typename... Types>
void print(T var1, Types... var2) {
  cout << var1 << ' ';
  print(var2...);
}
template <typename T>
void print(vector<T> &var1) {
  for (auto &i : var1) cout << i << ' ';
  cout << '\n';
}
template <typename T>
void print(T start, T end) {
  for (; start != end; start++) cout << *start << ' ';
  cout << '\n';
}
template <typename T>
void print(pair<T, T> &var1) {
  cout << var1.first << ' ' << var1.second << ' ';
  cout << '\n';
}
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename... Types>
void read(T &var1, Types &...var2) {
  cin >> var1;
  read(var2...);
}
template <typename T>
void read(vector<T> &var1) {
  for (auto &i : var1) read(i);
}
template <typename T>
void read(T start, T end) {
  for (; start != end; start++) cin >> *start;
}
template <typename T>
void read(pair<T, T> &var1) {
  cin >> var1.first >> var1.second;
}
template <typename T>
T Max(T x) {
  return x;
}
template <typename T, typename... Types>
T Max(T var1, Types... var2) {
  return var1 >= Max(var2...) ? var1 : Max(var2...);
}
template <typename T>
T Min(T x) {
  return x;
}
template <typename T, typename... Types>
T Min(T var1, Types... var2) {
  return var1 <= Min(var2...) ? var1 : Min(var2...);
}
long long INF = 1000000000000000005LL;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}
void solve() {
  long long l, r, m, a, b, c, n;
  read(l, r, m);
  long long val = r - l;
  for (long long i = l; i < r + 1; i++) {
    long long rem = m % i;
    if (rem <= val || (i - rem) <= val) {
      long long xy = (m - rem) / i;
      long long yz = (m + i - rem) / i;
      if ((xy > 0 && rem <= val)) {
        a = i;
        c = l;
        b = c + rem;
        print(a, b, c);
        return;
      }
      if (yz > 0 && (i - rem) <= val) {
        a = i;
        b = l;
        c = b + i - rem;
        print(a, b, c);
        return;
      }
    }
  }
}

```