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:

<image>

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.
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nlargest, nsmallest, _heapify_max, _heapreplace_max
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as c
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from fractions import Fraction
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


d = {"Tetrahedron": 4, "Cube": 6, "Octahedron": 8, "Dodecahedron": 12, "Icosahedron": 20}
answer = 0
for _ in range(int(data())):
    answer += d[data()]
out(answer)
