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;
int mx, mn, w, ans1, ans2, x, y, c;
map<double, int> m1, m2;
double qwq;
inline int f(int x) {
  int s = 0;
  while (x) {
    s = s * 10 + x % 10;
    x /= 10;
  }
  return s;
}
signed main() {
  scanf("%d%d%d", &mx, &mn, &w);
  x = mx;
  for (int i = 1; i <= mx; ++i) {
    qwq = 1.0 * i / f(i);
    ++m1[qwq];
  }
  while (x > 0 && y <= mn) {
    if (c < w && y < mn) {
      qwq = 1.0 * f(++y) / y;
      c += m1[qwq];
      ++m2[qwq];
    } else if (c >= w) {
      if (ans1 * ans2 > x * y || !ans1) ans1 = x, ans2 = y;
      qwq = 1.0 * x / f(x);
      c -= m2[qwq];
      --m1[qwq];
      --x;
    } else
      y = mn + 1;
  }
  if (ans1)
    printf("%d %d", ans1, ans2);
  else
    printf("-1");
  return 0;
}