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 sys
import math
import operator as op
from functools import reduce
from bisect import *
from collections import OrderedDict 
from collections import deque

## I/O ##
def read_line():
	return sys.stdin.readline()[:-1]
 
def read_int():
	return int(sys.stdin.readline())
	
def read_int_line():
	return [int(v) for v in sys.stdin.readline().split()]

def read_float_line():
	return [float(v) for v in sys.stdin.readline().split()]
## ##

## ncr ##
def ncr(n, r):
	r = min(r, n-r)
	numer = reduce(op.mul, range(n, n-r, -1), 1)
	denom = reduce(op.mul, range(1, r+1), 1)
	return numer / denom
## ##

## Fenwick Tree ##
# N = int(1e6)+10
# fenwick = [0]*N

# def add(i,x,n):
# 	while i<n:
# 		fenwick[i]+=x
# 		i = i|(i+1)

# def fsum(i,n):
# 	res = 0
# 	while i>=0:
# 		res += fenwick[i]
# 		i = (i&(i+1))-1
# 	return res

# def fsum_range(i,j,n):
# 	return fsum(j,n)-fsum(i-1,n)
## ##	

## binary search ##
def index(a, x):
    'Locate the leftmost value exactly equal to x'
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    return -1

def find_lt(a, x):
    'Find rightmost value less than x'
    i = bisect_left(a, x)
    if i:
        return i-1
    return -1

def find_le(a, x):
    'Find rightmost value less than or equal to x'
    i = bisect_right(a, x)
    if i:
        return i-1
    return -1

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return i
    return -1

def find_ge(a, x):
    'Find leftmost item greater than or equal to x'
    i = bisect_left(a, x)
    if i != len(a):
        return i
    return -1

## ##

## Main ##

mod = int(1e9)+7
mod2 = 998244353

t = read_int()
# t = 1
for i in range(t):
	a,b,n = read_int_line()
	a1 = a
	b1 = b
	cnt = 0
	f = 1
	
	while a<=n and b<=n:
		if f==1:
			b += a
			f *= -1
		else:
			a+=b
			f *= -1
		cnt+=1

	cnt1 = 0
	f1 = 1
	while a1<=n and b1<=n:
		if f1==-1:
			b1 += a1
			f1 *= -1
		else:
			a1+=b1
			f1 *= -1
		cnt1+=1

	print(min(cnt,cnt1))
