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!
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 *s*<sub class="lower-index">*i*</sub> — 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.
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.
4
Icosahedron
Cube
Tetrahedron
Dodecahedron
42
3
Dodecahedron
Octahedron
Octahedron
28
25
Icosahedron
Octahedron
Tetrahedron
Dodecahedron
Cube
Icosahedron
Octahedron
Cube
Tetrahedron
Icosahedron
Icosahedron
Tetrahedron
Octahedron
Dodecahedron
Icosahedron
Octahedron
Icosahedron
Tetrahedron
Dodecahedron
Tetrahedron
Octahedron
Cube
Cube
Dodecahedron
Tetrahedron
256
1
Tetrahedron
4
1
Cube
6
1
Octahedron
8
1
Dodecahedron
12
1
Icosahedron
20
28
Octahedron
Dodecahedron
Octahedron
Octahedron
Dodecahedron
Icosahedron
Icosahedron
Dodecahedron
Dodecahedron
Dodecahedron
Cube
Dodecahedron
Cube
Tetrahedron
Cube
Cube
Tetrahedron
Dodecahedron
Dodecahedron
Dodecahedron
Icosahedron
Icosahedron
Dodecahedron
Icosahedron
Dodecahedron
Dodecahedron
Icosahedron
Icosahedron
340
# Author :  //- kibrom Hailu -\\

from sys import stdin,stdout
from collections import Counter,defaultdict , deque 
from bisect import bisect , bisect_left ,bisect_right 
from itertools import accumulate 
from heapq import heapify , heappop , heappush , heappushpop , heapreplace

def I(): return int(stdin.readline())
def II(): return map(int, stdin.readline().split())
def IL(): return list(map(int, stdin.readline().split()))
def SIL(): return sorted(map(int, stdin.readline().split()))



def solve():
    
    # write your code here 
    n = I()

    def polygon_to_side(name:str):
        if name == "Tetrahedron":
            return 4
        elif name == "Cube":
            return 6 
        elif name == "Octahedron":
            return 8 
        elif name == "Dodecahedron":
            return 12 
        else:
            return 20
    
    total = 0 
    for _ in range(n):
        Current_name = input() 
        total += polygon_to_side(Current_name)
    
    return total 

    
    
    
    
    
    
    
    
print(solve())
