#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
bool desc(int a, int b) { return (a > b); }
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long a, b, n;
    cin >> a >> b >> n;
    int cnt = 0;
    if (a < b) {
      while (1) {
        if (cnt % 2 == 0) {
          a += b;
          cnt++;
        } else {
          b += a;
          cnt++;
        }
        if (a > n || b > n) {
          break;
        }
      }
    } else {
      long long t = a;
      a = b;
      b = t;
      while (1) {
        if (cnt % 2 == 0) {
          a += b;
          cnt++;
        } else {
          b += a;
          cnt++;
        }
        if (a > n || b > n) {
          break;
        }
      }
    }
    cout << cnt << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int INF = 1e9 + 7;
bool desc(int a, int b) { return (a > b); }
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long a, b, n;
    cin >> a >> b >> n;
    int cnt = 0;
    if (a < b) {
      while (1) {
        if (cnt % 2 == 0) {
          a += b;
          cnt++;
        } else {
          b += a;
          cnt++;
        }
        if (a > n || b > n) {
          break;
        }
      }
    } else {
      long long t = a;
      a = b;
      b = t;
      while (1) {
        if (cnt % 2 == 0) {
          a += b;
          cnt++;
        } else {
          b += a;
          cnt++;
        }
        if (a > n || b > n) {
          break;
        }
      }
    }
    cout << cnt << endl;
  }
  return 0;
}

```