Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.


-----Input-----

The single line contains integers n, m and k (1 ≤ n, m ≤ 5·10^5; 1 ≤ k ≤ n·m).


-----Output-----

Print the k-th largest number in a n × m multiplication table.


-----Examples-----
Input
2 2 2

Output
2

Input
2 3 4

Output
3

Input
1 10 5

Output
5



-----Note-----

A 2 × 3 multiplication table looks like this:

1 2 3

2 4 6

Difficulty: interview

Starter code:

Solutions:
def main():
    from math import sqrt
    m, n, k = list(map(int, input().split()))
    if n < m:
        n, m = m, n
    lo, hi = 1, k + 1
    while lo + 1 < hi:
        mid = (lo + hi) // 2
        t = mid - 1
        v = min(int(sqrt(t)), m)
        tn, tm = (t - 1) // m, t // n
        vv = [t // i for i in range(tm + 1, v + 1)]
        if t // n * (n + m) + sum(vv) * 2 + max(min((tn - tm), len(vv)) * m, 0) - v * v - sum(
                vv[:max(min(tn - tm, len(vv)), 0)]) < k:
            lo = mid
        else:
            hi = mid
    print(lo)


def __starting_point():
    main()

__starting_point()

from sys import stdin

n, m, k = [int(x) for x in stdin.readline().split()]
be, en = 1, k + 1

while be < en:
    mid = (be + en + 1) >> 1
    be1, cur = (mid + m - 1) // m, 0
    for i in range(1, be1):
        cur += m

    for i in range(be1, n + 1):
        cur += (mid - 1) // i

    if cur <= k - 1:
        be = mid
    else:
        en = mid - 1

print(be)


def works(X,N,M,K):
    #in each row, how many numbers are < X
    res = 0
    n = 1
    div = X/M
    while n < div:
        res += M
        n += 1
    while n < N+1:
        res += (X-1)//n
        n += 1
    return res

def solve():
    N, M, K = [int(s) for s in input().split()]
    left = 1
    right = K+1
    #we want the smallest smallest such that there are AT LEAST K-1 smaller numbers
    while right - left > 1:
        middle = (left+right)//2
        if works(middle,N,M,K) < K:
            left = middle
        else:
            right = middle
    #if there are exactly K-1 elements less than right, then this is our answer
    return left

#for _ in range(getInt()):    
print(solve())