The package for this problem was not updated by the problem writer or Codeforces administration after we've upgraded the judging servers. To adjust the time limit constraint, a solution execution time will be multiplied by 2. For example, if your solution works for 400 ms on judging servers, then the value 800 ms will be displayed and used to determine the verdict.

Codeforces Round 122 (Div. 1) |
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Finished |

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brute force

math

matrices

*2900

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E. Fibonacci Number

time limit per test

3 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputJohn Doe has a list of all Fibonacci numbers modulo 10^{13}. This list is infinite, it starts with numbers 0 and 1. Each number in the list, apart from the first two, is a sum of previous two modulo 10^{13}. That is, John's list is made from the Fibonacci numbers' list by replacing each number there by the remainder when divided by 10^{13}.

John got interested in number *f* (0 ≤ *f* < 10^{13}) and now wants to find its first occurrence in the list given above. Help John and find the number of the first occurence of number *f* in the list or otherwise state that number *f* does not occur in the list.

The numeration in John's list starts from zero. There, the 0-th position is the number 0, the 1-st position is the number 1, the 2-nd position is the number 1, the 3-rd position is the number 2, the 4-th position is the number 3 and so on. Thus, the beginning of the list looks like this: 0, 1, 1, 2, 3, 5, 8, 13, 21, ...

Input

The first line contains the single integer *f* (0 ≤ *f* < 10^{13}) — the number, which position in the list we should find.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print a single number — the number of the first occurrence of the given number in John's list. If this number doesn't occur in John's list, print -1.

Examples

Input

13

Output

7

Input

377

Output

14

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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