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;
map<pair<int, int>, int> A, B;
int maxx, maxy;
long long w, ansx = 100001, ansy = 100001;
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int rev(int x) {
  int tmp = 0;
  while (x) {
    tmp = tmp * 10 + x % 10;
    x /= 10;
  }
  return tmp;
}
int main() {
  scanf("%d%d%d", &maxx, &maxy, &w);
  for (int j = 1; j <= maxy; j++) {
    int g = gcd(rev(j), j), revj = rev(j);
    B[make_pair(revj / g, j / g)]++;
  }
  long long cnt = 0;
  int j = maxy, i;
  for (i = 1; i <= maxx; i++) {
    int g = gcd(rev(i), i), revi = rev(i);
    pair<int, int> tmp;
    tmp = make_pair(i / g, revi / g);
    A[tmp]++;
    cnt += B[tmp];
    while (1) {
      int g = gcd(rev(j), j), revj = rev(j);
      pair<int, int> tmp;
      tmp = make_pair(revj / g, j / g);
      if (cnt - A[tmp] < w) break;
      cnt -= A[tmp];
      B[tmp]--;
      j--;
    }
    if (cnt >= w)
      if ((long long)i * j < ansx * ansy ||
          (long long)i * j == ansx * ansy && i < ansx)
        ansx = i, ansy = j;
  }
  if (ansx <= maxx)
    printf("%I64d %I64d\n", ansx, ansy);
  else
    printf("-1");
  return 0;
}