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 x) {
  int y = 0;
  while (x) {
    y = y * 10 + x % 10;
    x /= 10;
  }
  return y;
}
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
pair<int, int> fx(int x) {
  int g = gcd(x, rev(x));
  return make_pair(x / g, rev(x) / g);
}
pair<int, int> fy(int x) {
  int g = gcd(x, rev(x));
  return make_pair(rev(x) / g, x / g);
}
map<pair<int, int>, int> mx;
map<pair<int, int>, int> my;
int main() {
  int maxx, maxy, w;
  scanf("%d %d %d", &maxx, &maxy, &w);
  mx.clear();
  my.clear();
  int ansx, ansy;
  ansx = ansy = -1;
  for (int y = 1; y <= maxy; y++) my[fy(y)]++;
  int total = 0;
  int y = maxy;
  for (int x = 1; x <= maxx; x++) {
    total += my[fx(x)];
    mx[fx(x)]++;
    while (y > 1 && total >= w) {
      int add = mx[fy(y)];
      if (total - add < w) break;
      my[fy(y)]--;
      total -= add;
      y--;
    }
    if (total >= w) {
      if (ansx == -1) {
        ansx = x;
        ansy = y;
      } else if (x * y < ansx * ansy) {
        ansx = x;
        ansy = y;
      }
    }
  }
  if (ansx == -1)
    printf("-1\n");
  else
    printf("%d %d\n", ansx, ansy);
  return 0;
}