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;
const int MAXN = 100000;
long long rev[MAXN];
struct cmp {
  bool operator()(const int &a, const int &b) const {
    return a * rev[b] < b * rev[a];
  }
};
map<int, int, cmp> map1, map2;
int X, Y, W;
int reverse(int x) {
  int ret = 0;
  while (x) {
    int dig = x % 10;
    x /= 10;
    ret = 10 * ret + dig;
  }
  return ret;
}
int main() {
  scanf("%d %d %d", &X, &Y, &W);
  for (int i = 0; i < MAXN; ++i) {
    rev[i] = reverse(i);
    if (reverse(rev[i]) != i) rev[i] = 0;
  }
  int a = 0;
  int b = 0;
  long long cnt = 0;
  while (b + 1 < MAXN) {
    ++b;
    ++map2[b];
  }
  int ansx = -1, ansy;
  long long best = -1;
  while (a < MAXN && b > 0) {
    if (cnt < W) {
      ++a;
      if (rev[a]) {
        cnt += map2[rev[a]];
      }
      ++map1[a];
    } else {
      if (a <= X && b <= Y) {
        long long prod = a;
        prod *= b;
        if ((best == -1) || prod < best) {
          best = prod;
          ansx = a;
          ansy = b;
        }
      }
      if (rev[b]) {
        cnt -= map1[rev[b]];
      }
      --map2[b];
      --b;
    }
  }
  if (ansx != -1) {
    printf("%d %d\n", ansx, ansy);
  } else {
    printf("-1\n");
  }
  return 0;
}