There are n boys and m girls attending a theatre club. To set a play "The Big Bang Theory", they need to choose a group containing exactly t actors containing no less than 4 boys and no less than one girl. How many ways are there to choose a group? Of course, the variants that only differ in the composition of the troupe are considered different.

Perform all calculations in the 64-bit type: long long for С/С++, int64 for Delphi and long for Java.

Input

The only line of the input data contains three integers n, m, t (4 ≤ n ≤ 30, 1 ≤ m ≤ 30, 5 ≤ t ≤ n + m).

Output

Find the required number of ways.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.

Examples

Input

5 2 5


Output

10


Input

4 3 5


Output

3
import sys
from math import factorial

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition) :
  if condition :
    print "YES"
  else :
    print "NO"

def plist(a, s = ' ') :
  print s.join(map(str, a))

def rint() :
  return int(sys.stdin.readline())

def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print '\n'.join(map(lambda x: separator.join(x), field)) 

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x /= p
  return digits

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n / 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

n, m, t = rints()

def C(n, m) :
  if n < m :
    return 0
 
  return factorial(n) / factorial(m) / factorial(n - m)

all_variants = C(n + m, t)
bad_variants = C(n, t) + C(m, t) + C(m, t - 1) * C(n, 1) + C(m, t - 2) * C(n, 2) + C(m, t - 3) * C(n, 3)

print all_variants - bad_variants
