human
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
Using python can you solve the prior task?
assistant
#Here is Python program to print -1.

from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
 
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
 
def ii():  return int(input())
def si():  return input()
def mi():  return map(int,input().split(" "))
def msi(): return map(str,input().split(" "))
def li():  return list(mi())
 
def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
    
#from math import *

 
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
    
def checkPrime(n) : # Check Prime Number or not 
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
    return True
 
def read():
    sys.stdin = open('input.txt', 'r')  
    sys.stdout = open('output.txt', 'w') 
def accordian(s):
    idx1=s.find("[")
    if idx1==-1:
        return("-1")
    s=s[idx1+1:]
    
    idx2=s.find(":")
    if idx2==-1:
        return("-1")
    s=s[idx2+1:]
    
    s=s[::-1]
    
    idx3=s.find("]")
    if idx3==-1:
        return("-1")
    s=s[idx3+1:]
   
    idx4=s.find(":")
    if idx4==-1:
        return("-1")
    s=s[idx4+1:]
    
    count=s.count("|")
    count+=4
    return count
def main():
    s=si()
    res=accordian(s)
    print(res)
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size,...