Problem: In Walrusland public transport tickets are characterized by two integers: by the number of the series and by the number of the ticket in the series. Let the series number be represented by a and the ticket number — by b, then a ticket is described by the ordered pair of numbers (a, b). 

The walruses believe that a ticket is lucky if a * b = rev(a) * rev(b). The function rev(x) reverses a number written in the decimal system, at that the leading zeroes disappear. For example, rev(12343) = 34321, rev(1200) = 21.

The Public Transport Management Committee wants to release x series, each containing y tickets, so that at least w lucky tickets were released and the total number of released tickets (x * y) were minimum. The series are numbered from 1 to x inclusive. The tickets in each series are numbered from 1 to y inclusive. The Transport Committee cannot release more than maxx series and more than maxy tickets in one series.

Input

The first line contains three integers maxx, maxy, w (1 ≤ maxx, maxy ≤ 105, 1 ≤ w ≤ 107).

Output

Print on a single line two space-separated numbers, the x and the y. If there are several possible variants, print any of them. If such x and y do not exist, print a single number  - 1.

Examples

Input

2 2 1


Output

1 1

Input

132 10 35


Output

7 5

Input

5 18 1000


Output

-1


Input

48 132 235


Output

22 111

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double a[100010], b[100010], rev[100010];
multimap<double, int> xmap, ymap;
multimap<double, int>::iterator xit[100010], yit[100010];
int mx, my, w;
int main(int argc, char** argv) {
  cin >> mx >> my >> w;
  for (int i = 1; i <= max(mx, my); i++) {
    rev[i] = 0;
    int k = i;
    while (k) {
      rev[i] = rev[i] * 10 + k % 10;
      k /= 10;
    }
    a[i] = (double)i / rev[i];
    b[i] = (double)rev[i] / i;
    if (i <= mx) xit[i] = xmap.insert(make_pair(a[i], i));
  }
  int cnt = 0;
  int rx, ry, rt = -1;
  for (int i = 1, j = mx; i <= my; i++) {
    cnt += xmap.count(b[i]);
    yit[i] = ymap.insert(make_pair(b[i], i));
    while (cnt >= w) {
      if (rt == -1 || i * j < rt) {
        rt = i * j;
        rx = j;
        ry = i;
      }
      cnt -= ymap.count(a[j]);
      xmap.erase(xit[j]);
      j -= 1;
    }
  }
  if (rt == -1)
    cout << rt << endl;
  else
    cout << rx << " " << ry << endl;
  return (EXIT_SUCCESS);
}