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 rev(int a) {
  int b = 0;
  while (a) {
    b = b * 10 + a % 10;
    a /= 10;
  }
  return b;
}
int gcd(int a, int b) {
  while (b) {
    int t = a % b;
    a = b;
    b = t;
  }
  return a;
}
int main() {
  int mx, my, w;
  scanf("%d%d%d", &mx, &my, &w);
  multimap<pair<int, int>, pair<int, int> > cor;
  for (int i = 1; i <= my; i++) {
    int r = rev(i);
    int d = gcd(i, r);
    int a = i / d, b = r / d;
    cor.insert(make_pair(make_pair(a, b), make_pair(i, r)));
  }
  static vector<int> list[110000];
  int count = 0;
  for (int i = 1; i <= mx; i++) {
    int r = rev(i);
    int d = gcd(i, r);
    int a = i / d, b = r / d;
    pair<int, int> t(b, a);
    for (multimap<pair<int, int>, pair<int, int> >::iterator p =
             cor.lower_bound(t);
         p != cor.end() && p->first == t; p++) {
      list[i].push_back(p->second.first);
      count++;
    }
  }
  if (count < w) {
    puts("-1");
    return 0;
  }
  priority_queue<int> ids;
  int sy = my, sx = mx;
  for (int i = 1; i <= mx; i++) {
    for (vector<int>::iterator j = list[i].begin(); j != list[i].end(); j++) {
      ids.push(*j);
    }
    while ((int)ids.size() > w) ids.pop();
    if ((int)ids.size() >= w) {
      int y = i;
      int x = ids.top();
      if ((long long)sy * (long long)sx > (long long)y * (long long)x) {
        sy = y;
        sx = x;
      }
    }
  }
  printf("%d %d\n", sy, sx);
  return 0;
}