#include <bits/stdc++.h>
using namespace std;
const long long INF64 = 1e18;
clock_t time_p = clock();
void time() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
void doc() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long n, a, b;
long long yo(long long a, long long b) {
  if (a > n and b > n) return 0;
  if (a < b) swap(a, b);
  return 1 + yo(a, a + b);
}
void solve() {
  long long ans = 0;
  cin >> a >> b >> n;
  if (a < b) swap(a, b);
  cout << yo(a, b) - 1 << '\n';
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your task is to create a cpp solution to the following problem:

Leo has developed a new programming language C+=. In C+=, integer variables can only be changed with a "+=" operation that adds the right-hand side value to the left-hand side variable. For example, performing "a += b" when a = 2, b = 3 changes the value of a to 5 (the value of b does not change).

In a prototype program Leo has two integer variables a and b, initialized with some positive values. He can perform any number of operations "a += b" or "b += a". Leo wants to test handling large integers, so he wants to make the value of either a or b strictly greater than a given value n. What is the smallest number of operations he has to perform?

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.

Each of the following T lines describes a single test case, and contains three integers a, b, n (1 ≤ a, b ≤ n ≤ 10^9) — initial values of a and b, and the value one of the variables has to exceed, respectively.

Output

For each test case print a single integer — the smallest number of operations needed. Separate answers with line breaks.

Example

Input


2
1 2 3
5 4 100


Output


2
7

Note

In the first case we cannot make a variable exceed 3 in one operation. One way of achieving this in two operations is to perform "b += a" twice.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF64 = 1e18;
clock_t time_p = clock();
void time() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
void doc() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long n, a, b;
long long yo(long long a, long long b) {
  if (a > n and b > n) return 0;
  if (a < b) swap(a, b);
  return 1 + yo(a, a + b);
}
void solve() {
  long long ans = 0;
  cin >> a >> b >> n;
  if (a < b) swap(a, b);
  cout << yo(a, b) - 1 << '\n';
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) solve();
}

```