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 inf = 1e9;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 100;
map<double, long long> mp1, mp2;
long long rev(long long x) {
  long long ans = 0;
  while (x) {
    ans = ans * 10 + x % 10;
    x /= 10;
  }
  return ans;
}
int main() {
  long long w, maxy, maxx, cnt = 0;
  scanf("%lld%lld%lld", &maxx, &maxy, &w);
  for (long long i = 1; i <= maxx; i++) mp1[1.0 * i / rev(i)]++;
  for (long long i = 1; i <= maxy; i++) cnt += mp1[1.0 * rev(i) / i];
  if (cnt < w) {
    cout << -1 << endl;
    return 0;
  }
  long long tx = maxx, ty = 1, ax = maxx, ay = maxy;
  cnt = mp1[1.0];
  mp2[1.0]++;
  while (tx >= 1 && ty <= maxy) {
    if (cnt >= w) {
      if (tx * ty < ax * ay) ax = tx, ay = ty;
      double tmp = 1.0 * tx / rev(tx);
      cnt -= mp2[tmp];
      mp1[tmp]--;
      tx--;
    } else {
      ty++;
      double tmp = 1.0 * rev(ty) / ty;
      cnt += mp1[tmp];
      mp2[tmp]++;
    }
  }
  printf("%lld %lld\n", ax, ay);
  return 0;
}