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;
map<double, long long> a1, a2;
long long maxx, maxy, w, cnt, l, r = 1, x, y;
double tmp;
inline long long qpow(register long long x) {
  long long ans = 0;
  while (x) {
    ans = ans * 10 + x % 10;
    x /= 10;
  }
  return ans;
}
int main() {
  scanf("%lld%lld%lld", &maxx, &maxy, &w);
  for (register int i = 1; i <= maxx; i++) a1[1.0 * i / qpow(i)]++;
  for (register int i = 1; i <= maxy; i++) cnt += a1[1.0 * qpow(i) / i];
  if (cnt < w) return puts("-1"), 0;
  l = x = maxx;
  y = maxy;
  cnt = a1[1.0];
  a2[1.0]++;
  while (l >= 1 && r <= maxy) {
    if (cnt >= w) {
      if (l * r < x * y) x = l, y = r;
      tmp = (double)l / qpow(l);
      cnt -= a2[tmp];
      a1[tmp]--;
      l--;
    } else {
      r++;
      tmp = (double)qpow(r) / r;
      cnt += a1[tmp];
      a2[tmp]++;
    }
  }
  printf("%lld %lld\n", x, y);
  return 0;
}