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






n, m, t = map(lambda t : int(t), input().split())

comDict = {}
comDict[n] = {}
comDict[m] = {}

nBoys= (max(4, t - m), min(n, max(t - m, t - 1)))
nGirls= (max(1, t - n), min(m, max(t - n, t - 4)))
#print (nBoys, nGirls)


def binom(n, m):
    b = [0] * (n + 1)
    b[0] = 1
    for i in range(1, n + 1):
        b[i] = 1
        j = i - 1
        while j > 0:
            b[j] += b[j - 1]
            j -= 1
    return b[m]

def comb (n, k):
    if k in comDict[n]:
        return comDict[n][k]
    #val = 1.0
    #for i in range(k):
    #    val *= (n - i) / (i + 1) # is 0 based
    #comDict[n][k] = int(val)
    comDict[n][k] = binom(n, k)
    return comDict[n][k]

s = 0
#c = []
for nn in range(nBoys[0], nBoys[1] + 1, 1):
    s += comb(n, nn) * comb(m, nGirls[1] - nn + nBoys[0])
    #print ('b ', nn, ' f ', t - nn)
    #c.append(('C(' + str(n) + ',' + str(nn) + ')*C(' + str(m) + ',' + str(nGirls[1] - nn + nBoys[0]) + ') +'))


# this is interestin https://en.wikipedia.org/wiki/Pascal%27s_triangle
# will go with the product https://wikimedia.org/api/rest_v1/media/math/render/svg/d4145e0326f57f563b59c943642928342a5a6b18

#print(" ".join(map(str,c)))
#print (comDict)
print (int(s))
