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.io.*;
 
public class Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0) {
			input=br.readLine().trim().split(" ");
			int a=Integer.parseInt(input[0]);
			int b=Integer.parseInt(input[1]);
			int n=Integer.parseInt(input[2]);
			
			out.println(minSteps(a,b,n));
		}
		
		out.flush();
		out.close();
	}	
	public static int minSteps(long a,long b,long n)
	{
		long tempA=a,tempB=b;
		int option1=0;
		while(a<=n && b<=n) {
			a+=b;
			option1++;
			if(a>n)
			{
				break;
			}
			b+=a;
			option1++;
		}
		
		a=tempA;
		b=tempB;
		int option2=0;
		while(a<=n && b<=n) {
			b+=a;
			option2++;
			if(b>n)
			{
				break;
			}
			a+=b;
			option2++;
		}
		int ans=Math.min(option2, option1);
		return ans;
	}
}