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.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class q2
{
	public static void main (String[] args) throws java.lang.Exception
	{
        // your code goes here
        Scanner sc = new Scanner(System.in);
        int t= Integer.parseInt(sc.next());
        while(t-->0)
        {
            long a= Long.parseLong(sc.next());
            long b= Long.parseLong(sc.next());
            long n= Long.parseLong(sc.next());
            long max=0;long min=0;
            long count=0;
            if(a>n || b>n)
            {
                System.out.println(0);
                continue;
            }
            if(a+b>n)
            {
                System.out.println(1);
                continue;
            }
            while(max<=n)
            {
                max=a+b;
                //b=a;
                long hj=Math.max(a,b);
                a=hj;
                b=max;
                count++;

            }
            System.out.println(count);
        }
    }
}