Problem: An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.

For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions. 

You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?

Input

The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.

Output

If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.

Examples

Input


|[a:b:|]


Output


4


Input


|]:[|:]


Output


-1

Solution (in PYTHON3):

Solution:
"""
  ____          _      _____
 / ___|___   __| | ___|  ___|__  _ __ ___ ___  ___
| |   / _ \ / _` |/ _ \ |_ / _ \| '__/ __/ _ \/ __|
| |__| (_) | (_| |  __/  _| (_) | | | (_|  __/\__ \
 \____\___/ \__,_|\___|_|  \___/|_|  \___\___||___/

"""
"""
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"""

import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt
from bisect import bisect_right, bisect_left

#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n - 1)


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  # or / in Python 2


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return len(set(factors))


def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))


def factors(n):
    return list(set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

MOD = 10**9 + 7
T = 1
#T = int(stdin.readline())
for _ in range(T):
    #n, k = list(map(int, stdin.readline().split()))
    # s1 = list(stdin.readline().strip('\n'))
    s = str(stdin.readline().strip('\n'))
    #n = int(stdin.readline())
    #l, r, d = list(map(int, stdin.readline().split()))
    #a = list(map(int, stdin.readline().split()))
    fb, lb = -1, -1
    n = len(s)
    for i in range(n):
        if s[i] == '[':
            fb = i
            break
    for i in range(n):
        if s[n - i - 1] == ']':
            lb = n - i - 1
            break
    if (fb > lb) or (fb == -1) or (lb == -1):
        print(-1)
        continue
    fc = -1
    lc = -1

    for i in range(fb, lb + 1):
        if s[i] == ':':
            fc = i
            break
    for i in range(lb, fb - 1, -1):
        if s[i] == ':':
            lc = i
            break

    if (fc == -1) or (lc == -1) or (fc == lc):
        print(-1)
        continue
    l = 0
    for i in range(fc, lc + 1):
        if s[i] == '|':
            l += 1
    print(2 + 2 + l)