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
input = sys.stdin.readline

'''

'''

def fast_factorial(mod=None):
    """
    Faster than naive factorial, uses memoization.
    """
    res = [1,1]
    size = 2

    def ff(k):
        nonlocal size

        while k >= size:
            res.append(res[-1] * size)
            size += 1  
        return res[k]
    
    def ff_mod(k):
        nonlocal size

        while k >= size:
            res.append((res[-1] * (size % mod)) % mod)
            size += 1
        
        return res[k]

    if mod == None:
        return ff
    else:
        return ff_mod

def extended_gcd(a, b):
    """returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
    s, old_s = 0, 1
    r, old_r = b, a
    while r:
        q = old_r // r
        old_r, r = r, old_r - q * r
        old_s, s = s, old_s - q * s
    return old_r, old_s, (old_r - old_s * a) // b if b else 0


def modinv(a, m):
    """returns the modular inverse of a w.r.t. to m, works when a and m are coprime"""
    g, x, _ = extended_gcd(a % m, m)
    return x % m if g == 1 else None

MOD = 10**9+7
ff = fast_factorial(MOD)
ffn = fast_factorial()

"""
def binom(n, k):
    return (ff(n) * modinv((ff(k) * ff(n-k) % MOD), MOD)) % MOD
"""

def binom(n, k):
    return (ffn(n) // (ffn(k)*(ffn(n-k)))) 

n, m, t = map(int, input().split())
res = 0

for boys in range(4, n+1):
    if 1 <= t-boys <= m:
        res += (binom(n, boys) * binom(m, t-boys)) #% MOD
        #res %= MOD

print(res)



