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;
long long maxx, maxy, W, cnt, ansx, ansy;
map<double, long long> map1, map2;
inline int rev(int x) {
  int ans = 0;
  for (; x; x /= 10) (ans *= 10) += x % 10;
  return ans;
}
int main() {
  scanf("%lld%lld%lld", &maxx, &maxy, &W);
  for (int i = (1); i <= (maxx); ++i) ++map1[1. * i / rev(i)];
  for (int i = (1); i <= (maxy); ++i) cnt += map1[1. * rev(i) / i];
  if (cnt < W) return puts("-1"), 0;
  long long tx = maxx, ty = 1LL;
  long long ansx = maxx, ansy = maxy;
  cnt = map1[1.], ++map2[1.];
  while (tx && ty ^ maxy) {
    if (cnt >= W) {
      if (tx * ty < ansx * ansy) ansx = tx, ansy = ty;
      double quo = 1. * tx / rev(tx);
      --tx;
      cnt -= map2[quo], --map1[quo];
    } else {
      ++ty;
      double quo = 1. * rev(ty) / ty;
      cnt += map1[quo], ++map2[quo];
    }
  }
  printf("%lld %lld\n", ansx, ansy);
  return 0;
}