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;
long long anss = 1e15, ansx, ansy;
map<double, long long> mp1, mp2;
double x[500005], y[500005];
long long work(long long x) {
  long long ans = 0;
  while (x) {
    ans = ans * 10 + x % 10;
    x /= 10;
  }
  return ans;
}
signed main() {
  long long xx, yy, w, ans = 0;
  scanf("%lld%lld%lld", &xx, &yy, &w);
  for (long long i = 1; i <= xx; i++) {
    x[i] = i;
    x[i] /= work(i);
    mp1[x[i]]++;
  }
  for (long long i = 1; i <= yy; i++) {
    y[i] = work(i);
    y[i] /= i;
  }
  long long xxx = xx, yyy = 1;
  ans += mp1[y[1]];
  mp2[y[1]]++;
  while (xxx >= 1 && yyy <= yy) {
    if (ans >= w) {
      if (anss > xxx * yyy) {
        ansx = xxx;
        ansy = yyy;
        anss = xxx * yyy;
      }
      ans -= mp2[x[xxx]];
      mp1[x[xxx]]--;
      xxx--;
    } else {
      yyy++;
      ans += mp1[y[yyy]];
      mp2[y[yyy]]++;
    }
  }
  if (anss == 1e15)
    printf("-1\n");
  else
    printf("%lld %lld\n", ansx, ansy);
  return 0;
}