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.
import java.util.Scanner;

public class Lista1_C {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = Integer.parseInt(sc.nextLine());

        for (int i = 0; i < T; i++) {
            String[] input = sc.nextLine().split(" ");
            int first = Integer.parseInt(input[0]) < Integer.parseInt(input[1]) ? Integer.parseInt(input[0]) : Integer.parseInt(input[1]);
            int second = Integer.parseInt(input[0]) > Integer.parseInt(input[1]) ? Integer.parseInt(input[0]) : Integer.parseInt(input[1]);
            int finalN = Integer.parseInt(input[2]);

            System.out.println(fib(first, second, finalN, 0));
        }
        sc.close();
    }

    public static int fib(int first, int second, int finalN, int count) {
        if (first + second > finalN) {
            count++;
            return count;
        }
        if (count % 2 == 0) {
            first += second;
        } else {
            second += first;
        }
        count++;
        return fib(first, second, finalN, count);
    }
}
