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;
double aux(long long x) {
  long long ans = 0;
  while (x) ans = ans * 10 + x % 10, x /= 10;
  return (double)ans;
}
map<double, long long> mp1, mp2;
double f[200005];
long long n, m, w;
signed main() {
  scanf("%I64d%I64d%I64d", &n, &m, &w);
  for (long long i = 1; i <= n; i++) f[i] = i / aux(i), mp1[f[i]]++;
  long long tot = 0;
  for (long long i = 1; i <= m; i++) tot += mp1[aux(i) / i];
  if (tot < w) return puts("-1"), 0;
  long long xx = n, yy = 1;
  long long ansx, ansy, plk = 1e18;
  long long sum = mp1[1.0];
  mp2[1.0]++;
  while (xx >= 1 && yy <= m) {
    if (sum >= w) {
      if (xx * yy < plk) plk = xx * yy, ansx = xx, ansy = yy;
      double pl = 1.0 * xx / aux(xx);
      mp1[pl]--;
      sum -= mp2[pl];
      xx--;
    } else {
      yy++;
      double pl = aux(yy) / yy;
      sum += mp1[pl];
      mp2[pl]++;
    }
  }
  cout << ansx << " " << ansy;
  return 0;
}