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.
/*==============================================
Name        : Shadman Shariar                 ||                    
Email       : shadman.shariar@northsouth.edu  ||
University  : North South University (NSU)    ||
Facebook    : shadman.shahriar.007            ||
==============================================*/ 

import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		
		Scanner input = new Scanner (System.in);
		
		int t = input.nextInt();
		
		for (int i = 0; i < t; i++) {
			int c =0;
			int a = input.nextInt();
			int b = input.nextInt();
		    int n = input.nextInt();	
		    while (Math.max(a,b)<=n)
	        {
	            if (a<b)    a+=b;
	            else        b+=a;
	            c++;
	        }
		System.out.println(c);	
		}
		
		
		input.close();
	}
}