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, sum;
map<long double, long long> mp, Mp;
inline long long rev(long long x) {
  long long sum = 0;
  while (x) {
    sum = sum * 10 + x % 10;
    x /= 10;
  }
  return sum;
}
int main() {
  cin >> maxx >> maxy >> w;
  for (long long i = 1; i <= maxx; i++) mp[i * 1.0 / rev(i)]++;
  for (long long i = 1; i <= maxy; i++) sum += mp[rev(i) * 1.0 / i];
  if (sum < w) {
    printf("-1\n");
    return 0;
  }
  long long ansx = maxx, ansy = maxy, X = maxx, Y = 1;
  sum = mp[1];
  Mp[1]++;
  while (X >= 1 && Y <= maxy) {
    if (sum >= w) {
      if (X * Y < ansx * ansy) {
        ansx = X;
        ansy = Y;
      }
      long double x = X * 1.0 / rev(X);
      sum -= Mp[x];
      mp[x]--;
      X--;
    } else {
      Y++;
      long double x = rev(Y) * 1.0 / Y;
      sum += mp[x];
      Mp[x]++;
    }
  }
  printf("%lld %lld\n", ansx, ansy);
  return 0;
}