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 LeoC {

    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int x= sc.nextInt();
        int a[][]= new int[x][3];//declaring an array?
        for(int i = 0;i<x; i++){
            for(int j=0;j<3;j++){
                a[i][j]=sc.nextInt();
//                System.out.print(a[i][j]+" ");
            }
//            System.out.println();
        }//inputs done
        int ans=0;
        for(int i=0;i<x;i++){
            ans=0;
            int max=a[i][2];
            int nMax= Math.max(a[i][0], a[i][1]);
            int min= Math.min(a[i][0], a[i][1]);

            while(nMax<=max){
//                System.out.println(nMax+" "+min);
                int t=nMax;
                nMax+=min;
                min=t;
//                int t=min;
//                min=nMax;
//                nMax=t;
                ans++;
            }
            System.out.println(ans);
        }


    }
}
