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.*; 
public class Main
{
    
	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t>0){
		    long a = sc.nextLong();
		    long b = sc.nextLong();
		    long n = sc.nextLong();
		    if(a>n || b>n){
                System.out.println(0);		        
		    }
            else if(a==n || b==n){
                System.out.println(1);
            }
            else{
                long maxOf = Math.max(a,b);
                long minOf = Math.min(a,b);
                long cnt = 0;
                long tminOf = 0;
                long tmaxOf = 0;
                while((Math.max(maxOf,minOf))<=n){
                    tmaxOf = Math.max(maxOf,minOf);
                    tminOf = Math.min(maxOf,minOf);
                    tminOf += tmaxOf;
                    maxOf = tmaxOf;
                    minOf = tminOf;
                    
                    
                    cnt++;
                }
    
                System.out.println(cnt);                
            }

		    t--;
		}
	}
}
