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 i, j, k, n, m, K, an1, an2;
map<double, int> Map1, Map2;
long long mi, sum;
inline int get(int x) {
  int s = 0;
  for (; x; x /= 10) s = s * 10 + x % 10;
  return s;
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= n; i++) Map1[1. * get(i) / i]++;
  mi = (long long)1e15;
  j = n;
  for (i = 1; i <= m; i++) {
    sum += Map1[1. * i / get(i)];
    Map2[1. * i / get(i)]++;
    for (; j > 1;) {
      int A = Map2[1. * get(j) / j];
      if (sum - A >= K) {
        sum -= A;
        Map1[1. * get(j) / j]--;
        j--;
      } else
        break;
    }
    if (sum >= K && 1ll * i * j < mi) mi = 1ll * i * j, an1 = j, an2 = i;
  }
  if (mi == 1e15)
    printf("-1\n");
  else
    printf("%d %d\n", an1, an2);
  return 0;
}