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 a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int rev(int n) {
  int ret = 0;
  while (n != 0) {
    ret = (10 * ret) + (n % 10);
    n /= 10;
  }
  return ret;
}
int main() {
  int mx, my, w;
  int i, j, k, t;
  int ax, ay;
  vector<int> v;
  map<pair<int, int>, vector<int> > mp;
  scanf("%d%d%d", &mx, &my, &w);
  for (i = 1; i <= my; i++) {
    j = rev(i);
    k = gcd(i, j);
    mp[make_pair(i / k, j / k)].push_back(i);
  }
  vector<int> vec(my + 1, 0);
  vector<vector<int> > vv;
  vv.push_back(v);
  for (i = 1; i <= mx; i++) {
    j = rev(i);
    k = gcd(i, j);
    v = mp[make_pair(j / k, i / k)];
    for (j = 0; j < v.size(); j++) vec[v[j]]++;
    vv.push_back(v);
  }
  long long ans = -1;
  for (i = mx, j = 0, k = 0; i > 0; i--) {
    if (k < w && j < my) {
      do {
        j++;
        k += vec[j];
      } while (k < w && j < my);
    }
    if (k >= w && (ans == -1 || ans > (long long)i * j)) {
      ans = (long long)i * j;
      ax = i;
      ay = j;
    }
    for (t = 0; t < vv[i].size(); t++) {
      if (vv[i][t] <= j) k--;
      vec[vv[i][t]]--;
    }
  }
  if (ans == -1)
    printf("-1\n");
  else
    printf("%d %d\n", ax, ay);
  return 0;
}