#include <bits/stdc++.h>
using namespace std;
int mx, mn, w, x, y, ansx, ansy, co = 0;
int rev(int k) {
  int p = 0;
  while (k > 0) {
    p = p * 10 + (k % 10);
    k /= 10;
  }
  return p;
}
int main() {
  map<double, int> m1;
  map<double, int> m2;
  double q;
  cin >> mx;
  cin >> mn;
  cin >> w;
  for (x = 1; x <= mx; x++) {
    q = (double)(x * 1.0 / rev(x));
    m1[q]++;
  }
  x = mx;
  y = 0;
  co = 0;
  while (y <= mn && x >= 1) {
    if (co < w) {
      y++;
      q = (double)((rev(y) * 1.0) / y);
      co += m1[q];
      m2[q]++;
    } else {
      if (ansx * ansy > x * y || ansx == 0) {
        ansx = x;
        ansy = y;
      }
      q = (double)((x * 1.0) / rev(x));
      co -= m2[q];
      m1[q]--;
      x--;
    }
  }
  if (ansx == 0)
    cout << -1;
  else
    cout << ansx << "  " << ansy;
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int mx, mn, w, x, y, ansx, ansy, co = 0;
int rev(int k) {
  int p = 0;
  while (k > 0) {
    p = p * 10 + (k % 10);
    k /= 10;
  }
  return p;
}
int main() {
  map<double, int> m1;
  map<double, int> m2;
  double q;
  cin >> mx;
  cin >> mn;
  cin >> w;
  for (x = 1; x <= mx; x++) {
    q = (double)(x * 1.0 / rev(x));
    m1[q]++;
  }
  x = mx;
  y = 0;
  co = 0;
  while (y <= mn && x >= 1) {
    if (co < w) {
      y++;
      q = (double)((rev(y) * 1.0) / y);
      co += m1[q];
      m2[q]++;
    } else {
      if (ansx * ansy > x * y || ansx == 0) {
        ansx = x;
        ansy = y;
      }
      q = (double)((x * 1.0) / rev(x));
      co -= m2[q];
      m1[q]--;
      x--;
    }
  }
  if (ansx == 0)
    cout << -1;
  else
    cout << ansx << "  " << ansy;
}

```