Problem:
Score: 500 points

Problem Statement

In the Kingdom of AtCoder, only banknotes are used as currency. There are 10^{100}+1 kinds of banknotes, with the values of 1, 10, 10^2, 10^3, \dots, 10^{(10^{100})}. You have come shopping at a mall and are now buying a takoyaki machine with a value of N. (Takoyaki is the name of a Japanese snack.)

To make the payment, you will choose some amount of money which is at least N and give it to the clerk. Then, the clerk gives you back the change, which is the amount of money you give minus N.

What will be the minimum possible number of total banknotes used by you and the clerk, when both choose the combination of banknotes to minimize this count?

Assume that you have sufficient numbers of banknotes, and so does the clerk.

Constraints

N is an integer between 1 and 10^{1,000,000} (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum possible number of total banknotes used by you and the clerk.

Sample Input 1

36

Sample Output 1

8

If you give four banknotes of value 10 each, and the clerk gives you back four banknotes of value 1 each, a total of eight banknotes are used.

The payment cannot be made with less than eight banknotes in total, so the answer is 8.

Sample Input 2

91

Sample Output 2

3

If you give two banknotes of value 100, 1, and the clerk gives you back one banknote of value 10, a total of three banknotes are used.

Sample Input 3

314159265358979323846264338327950288419716939937551058209749445923078164062862089986280348253421170

Sample Output 3

243

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys
read = sys.stdin.readline


# 下の桁から貪欲に見ていくと解ける
# 1の位が4以下ならそのまま紙幣で支払うのが得→自身をそのまま答えに採用 (枚数として加算)
# 6以上ならばお釣り狙いのほうが得→自身は0にセットして、上位桁を+1する (枚数として10-自身を加算)
# 5ちょうどのときは、お釣りもそのまま払っても枚数は変わらない。だけど555みたいに、上位桁に+1すると得になる可能性がある。
# 555→15枚, 1000-445→15枚
# つまり5が出たときは更に上位の桁を見て、4以下ならそのまま払うほうが得,5以上なら桁上りのほうが得

# 5ちょうどのときは？ #お釣りもそのまま払っても同じ。だけど上位 お釣り狙いだと上の桁が多くなって支払う金額が増える可能性がある？
# 5ピッタリのときに上位桁に+1すると悪くなる場合が存在する 455 とは 15にってしまう
# 5ピッタリのときはどうするのがよいのか？

N = '0' + read()[:-1]

N = N[::-1]
ans = 0
i = 0

lo = int(N[i])
while i < len(N) - 1:
    hi = int(N[i + 1])
    if lo == 5:  # 5のときは上の桁の大きさによってお釣りが得になるかそのまま払うのが良くなるか変化する
        ans += lo
        if hi < 5:
            pass
        else:
            hi += 1
    elif lo < 5:
        # loが5以外のとき (お釣りが良いか、そのまま払うのが良いか考えるだけ)
        ans += lo
    else:
        ans += 10 - lo
        hi += 1  # 上の桁が繰り上がる
    lo = hi  # 桁を一つずらす
    i += 1


print(ans)