#include <bits/stdc++.h>
int exceed(int a, int b, int n);
int main() {
  int T;
  scanf("%d", &T);
  int a, b, n;
  while (T--) {
    scanf("%d %d %d", &a, &b, &n);
    printf("%d\n", exceed(a, b, n));
  }
}
int exceed(int a, int b, int n) {
  int count = 0, cycle = 0;
  while (a <= n && b <= n) {
    if (a <= b && cycle == 0) {
      a += b;
      cycle = 1;
    } else if (b < a && cycle == 0) {
      b += a;
      cycle = 2;
    } else if (cycle == 2) {
      if (count % 2 == 0) {
        b += a;
      } else {
        a += b;
      }
    } else if (cycle == 1) {
      if (count % 2 == 0) {
        a += b;
      } else {
        b += a;
      }
    }
    count++;
  }
  return count;
}

### Prompt

Generate 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>
int exceed(int a, int b, int n);
int main() {
  int T;
  scanf("%d", &T);
  int a, b, n;
  while (T--) {
    scanf("%d %d %d", &a, &b, &n);
    printf("%d\n", exceed(a, b, n));
  }
}
int exceed(int a, int b, int n) {
  int count = 0, cycle = 0;
  while (a <= n && b <= n) {
    if (a <= b && cycle == 0) {
      a += b;
      cycle = 1;
    } else if (b < a && cycle == 0) {
      b += a;
      cycle = 2;
    } else if (cycle == 2) {
      if (count % 2 == 0) {
        b += a;
      } else {
        a += b;
      }
    } else if (cycle == 1) {
      if (count % 2 == 0) {
        a += b;
      } else {
        b += a;
      }
    }
    count++;
  }
  return count;
}

```