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 maxx, maxy, w;
map<pair<int, int>, int> cntx, cnty;
int mcd(int a, int b) {
  int tmp;
  while (b != 0) {
    tmp = b;
    b = a % b;
    a = tmp;
  }
  return a;
}
int rev(int n) {
  int ret = 0;
  while (n > 0) {
    ret = 10 * ret + (n % 10);
    n /= 10;
  }
  return ret;
}
pair<int, int> mpair(int x, int rx) {
  int d = mcd(x, rx);
  return pair<int, int>(rx / d, x / d);
}
int main() {
  cin >> maxx >> maxy >> w;
  long long bestx = maxx, besty = maxy;
  cntx.clear();
  cnty.clear();
  for (int x = 1; x <= maxx; x++) cntx[mpair(x, rev(x))]++;
  long long y = 1, x = maxx;
  int cur_count = cntx[mpair(rev(1), 1)];
  cnty[mpair(rev(1), 1)]++;
  while (y < maxy && cur_count < w) {
    y++;
    cur_count += cntx[mpair(rev(y), y)];
    cnty[mpair(y, rev(y))]++;
  }
  if (y == maxy && cur_count < w) {
    cout << -1 << endl;
    return 0;
  }
  while (x >= 1 && y <= maxy) {
    while (x >= 1 && cur_count - cnty[mpair(rev(x), x)] >= w) {
      cntx[mpair(x, rev(x))]--;
      cur_count -= cnty[mpair(rev(x), x)];
      x--;
    }
    if (x * y < bestx * besty) {
      bestx = x;
      besty = y;
    }
    y++;
    if (y <= maxy) {
      cur_count += cntx[mpair(rev(y), y)];
      cnty[mpair(y, rev(y))]++;
    }
  }
  cout << bestx << " " << besty << endl;
  return 0;
}