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 MA, MB, W;
int r1, r2;
int rev(int u) {
  int v = 0;
  for (; u; u /= 10) v = v * 10 + u % 10;
  return v;
}
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int main() {
  scanf("%d%d%d", &MA, &MB, &W);
  for (int i = 1; i <= MA; i++) {
    int u = i, v = rev(i);
    A[make_pair(u / gcd(u, v), v / gcd(u, v))]++;
  }
  r1 = r2 = -1;
  long long now = 0;
  for (int i = MA, j = 0; i; i--) {
    while (j < MB && now < W) {
      ++j;
      int u = rev(j), v = j;
      pair<int, int> tmp = make_pair(u / gcd(u, v), v / gcd(u, v));
      now += A[tmp];
      B[tmp]++;
    }
    if (now >= W) {
      if (r1 == -1 || (long long)r1 * r2 > (long long)i * j) r1 = i, r2 = j;
    }
    int u = i, v = rev(i);
    pair<int, int> tmp = make_pair(u / gcd(u, v), v / gcd(u, v));
    now -= B[tmp];
    A[tmp]--;
  }
  if (r1 == -1)
    printf("-1\n");
  else
    printf("%d %d\n", r1, r2);
}