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 in() {
  int a;
  scanf("%d", &a);
  return a;
}
int gcm(int a, int b) {
  if (b > a) return gcm(b, a);
  if (a % b == 0)
    return b;
  else
    return gcm(b, a % b);
}
int calc_rev(int x) {
  int ret = 0;
  while (x) {
    int a = x % 10;
    ret = ret * 10 + a;
    x /= 10;
  }
  return ret;
}
int rev[100010];
map<pair<int, int>, vector<int> > lucky;
pair<int, int> a_reva(int a) {
  int g = gcm(a, rev[a]);
  int b = a / g;
  int c = rev[a] / g;
  return make_pair(b, c);
}
pair<int, int> invert_pint(pair<int, int> t) {
  return make_pair(t.second, t.first);
}
int main() {
  for (int i = (1); i <= (100000); i++) {
    rev[i] = calc_rev(i);
  }
  for (int i = (1); i <= (100000); i++) {
    lucky[a_reva(i)].push_back(i);
  }
  pair<int, int> ans = make_pair(-1, -1);
  long long ans_fact = ((long long)(1001001001) * (1001001001));
  int maxx = in();
  int maxy = in();
  int w = in();
  int bar = maxy;
  int ltickets = 0;
  for (int x = (1); x <= (maxx); x++) {
    vector<int> hoge = lucky[invert_pint(a_reva(x))];
    int ind =
        distance(hoge.begin(), upper_bound(hoge.begin(), hoge.end(), bar));
    ltickets += ind;
    while (ltickets >= w) {
      if ((long long)x * bar < ans_fact) {
        ans = make_pair(x, bar);
        ans_fact = (long long)x * bar;
      }
      vector<int> fuga = lucky[invert_pint(a_reva(bar))];
      int ind2 =
          distance(fuga.begin(), upper_bound(fuga.begin(), fuga.end(), x));
      ltickets -= ind2;
      bar--;
    }
  }
  if (ans.first == -1) {
    puts("-1");
  } else {
    printf("%d %d\n", ans.first, ans.second);
  }
  return 0;
}