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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class Greater {

	public static void main(String[]args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pr = new PrintWriter(System.out);
	    int n = Integer.parseInt(br.readLine());
	    for(int i=0 ; i<n ; i++){
	    	StringTokenizer st = new StringTokenizer(br.readLine());
	    	int a = Integer.parseInt(st.nextToken());
	    	int b = Integer.parseInt(st.nextToken());
	    	int t = Integer.parseInt(st.nextToken());
	    	int c = 0;
	    	while(a<=t && b<=t){
	    		if(a<b)
	    			a+=b;
	    		else
	    			b+=a;
	    		c++;
	    	}
	    	pr.println(c);
	    }
	    pr.flush();
	}
}