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;
struct dat {
  long long x, y;
  bool operator<(dat const &T) const { return x * T.y < y * T.x; }
  bool operator==(dat const &T) const { return x * T.y == y * T.x; }
} T;
int rev(int x) {
  int R = 0, t;
  while (x > 0) {
    t = x % 10;
    R = R * 10 + t;
    x /= 10;
  }
  return R;
}
dat a[100005];
map<dat, int> X, Y;
int mx, my;
int ret = -1, rex, rey, par, n, t, pt;
int main() {
  scanf("%d%d%d", &mx, &my, &n);
  Y.clear();
  for (int i = 1; i <= my; i++) {
    a[i].x = i;
    a[i].y = rev(i);
    Y[a[i]] = Y[a[i]] + 1;
  }
  X.clear();
  par = 0;
  pt = my;
  for (int i = 1; i <= mx; i++) {
    T.x = rev(i);
    T.y = i;
    par += Y[T];
    X[T] = X[T] + 1;
    if (par < n) continue;
    while (pt >= 1) {
      T.x = a[pt].x;
      T.y = a[pt].y;
      t = X[T];
      if (par - t >= n) {
        par -= t;
        Y[a[pt]] = Y[a[pt]] - 1;
        pt--;
      } else
        break;
    }
    if (i * pt < ret || ret == -1) {
      ret = i * pt;
      rex = i;
      rey = pt;
    }
  }
  if (ret == -1)
    printf("-1\n");
  else
    printf("%d %d\n", rex, rey);
  return 0;
}