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 gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int rev(int x) {
  int ret = 0;
  while (x) {
    ret = ret * 10 + x % 10;
    x /= 10;
  }
  return ret;
}
map<pair<int, int>, int> m1, m2;
int ansx, ansy;
void check(int x, int y) {
  if (ansx == -1 || ((long long)ansx) * ansy > ((long long)x) * y)
    ansx = x, ansy = y;
}
int main() {
  int xmax, ymax, w;
  ansx = ansy = -1;
  scanf("%d%d%d", &xmax, &ymax, &w);
  int now = 0, x = 0, y = ymax;
  for (int i = 1, j, d; i <= ymax; i++) {
    j = rev(i);
    d = gcd(i, j);
    m2[make_pair(i / d, j / d)]++;
  }
  for (x = 1; x <= xmax; x++) {
    int i = x, j = rev(x), d = gcd(i, j);
    m1[make_pair(i / d, j / d)]++;
    now += m2[make_pair(j / d, i / d)];
    while (now >= w) {
      check(x, y);
      i = y, j = rev(y), d = gcd(i, j);
      m2[make_pair(i / d, j / d)]--;
      now -= m1[make_pair(j / d, i / d)];
      y--;
    }
  }
  if (ansx == -1)
    puts("-1");
  else
    printf("%d %d\n", ansx, ansy);
  return 0;
}