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.
Solve the task in PYTHON3.

=====
Can you provide a Python code to find this number!?
assistant
#Here is Python code to find this number!
.

import sys,os
from io import BytesIO, IOBase
import collections,itertools,bisect,heapq,math,string
# region fastio
 
BUFSIZE = 8192
 
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), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def main():
    n = int(input())
    a = 0
    for i in range(n):
        s = input()
        if s == 'Tetrahedron':
            b = 4
        elif s == 'Cube':
            b = 6
        elif s == 'Octahedron':
       ...