human
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: 

  * Tetrahedron. Tetrahedron has 4 triangular faces. 
  * Cube. Cube has 6 square faces. 
  * Octahedron. Octahedron has 8 triangular faces. 
  * Dodecahedron. Dodecahedron has 12 pentagonal faces. 
  * Icosahedron. Icosahedron has 20 triangular faces. 


All five kinds of polyhedrons are shown on the picture below:


Anton has a collection of n polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of polyhedrons in Anton's collection.

Each of the following n lines of the input contains a string si — the name of the i-th polyhedron in Anton's collection. The string can look like this:

  * "Tetrahedron" (without quotes), if the i-th polyhedron in Anton's collection is a tetrahedron. 
  * "Cube" (without quotes), if the i-th polyhedron in Anton's collection is a cube. 
  * "Octahedron" (without quotes), if the i-th polyhedron in Anton's collection is an octahedron. 
  * "Dodecahedron" (without quotes), if the i-th polyhedron in Anton's collection is a dodecahedron. 
  * "Icosahedron" (without quotes), if the i-th polyhedron in Anton's collection is an icosahedron. 

Output

Output one number — the total number of faces in all the polyhedrons in Anton's collection.

Examples

Input

4
Icosahedron
Cube
Tetrahedron
Dodecahedron

Output

42

Input

3
Dodecahedron
Octahedron
Octahedron

Output

28

Note

In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
Using python can you solve the prior task?
 ... Give me a Python program to find this number!?
assistant
#Here is Python program to find this number!
.

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 main():
    tot=0
    for _ in range(ii()):
        shape=si()
        if shape=="Tetrahedron":
            tot+=4
        elif shape=="Cube":
            tot+=6
        elif shape=="Octahedron":
            tot+=8
        elif shape=="Dodecahedron":
            tot+=12
        elif shape=="Icosahedron":
            tot+=20
    print(tot)

# 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, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2),...