Let's assume that we have a pair of numbers (*a*, *b*). We can get a new pair (*a* + *b*, *b*) or (*a*, *a* + *b*) from the given pair in a single step.

Let the initial pair of numbers be (1,1). Your task is to find number *k*, that is, the least number of steps needed to transform (1,1) into the pair where at least one number equals *n*.
The input contains the only integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">6</sup>).
5
3
1
0
2
1
3
2
4
3
10
5
1009
15
2009
17
7009
19
9009
20
19009
21
29009
22
12434
21
342342
28
342235
28
362235
28
762235
30
878235
30
978235
30
1000000
30
6
5
10000
20
999999
30
524287
29
777777
30
123756
26
976438
30
434563
28
345634
28
65457
24
123456
26
999997
30
98989
25
8
4
123455
26
990001
30
123141
26
998
16
453422
28
623423
29
89
9
24234
23
999879
30
345612
28
998756
30
999989
30
999998
30
999912
30
100000
25
def calc(n, i):
    # print(n, i)
    if i > n:
        n, i = i, n
    if n == i and i == 1:
        return 0
    if (n % 2 == 0 and i % 2 == 0) or i == 0:
        return 10000000
    else:
        return 1 + calc(n - i, i)

fibo = [0, 1, 2]
a, b, c = 1, 2, 0
for i in range(1, 1000):
    c = a + b
    fibo.append(c)
    b, a = c, b

n = int(input())
if n == 1:
    print(0)
else:
    for i in range(0, len(fibo)):
        if n > fibo[i]:
            continue
        else:
            print(calc(n, fibo[i - 1]))
            break