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 mul = 1e9;
long long n, m, w, ans, ansx, ansy;
map<long long, long long> a, b;
long long Rev(long long x) {
  long long res = 0;
  while (x) {
    res = res * 10 + x % 10;
    x /= 10;
  }
  return res;
}
signed main() {
  cin >> n >> m >> w;
  ans = n * m + 1;
  {
    long long i = 0, j = m, now = 0;
    for (long long o = 1; o <= m; ++o) {
      long long tmp = round(mul * Rev(o) / o);
      if (b.find(tmp) != b.end())
        ++b[tmp];
      else
        b[tmp] = 1;
    }
    while (i <= n && j >= 1) {
      if (now < w) {
        ++i;
        long long tmp = round(mul * i / Rev(i));
        if (a.find(tmp) != a.end())
          ++a[tmp];
        else
          a[tmp] = 1;
        if (b.find(tmp) != b.end()) now += b[tmp];
      } else {
        if (ans > i * j) {
          ans = i * j;
          ansx = i;
          ansy = j;
        }
        long long tmp = round(mul * Rev(j) / j);
        if (a.find(tmp) != a.end()) now -= a[tmp];
        --b[tmp];
        --j;
      }
    }
  }
  if (ans == n * m + 1) {
    puts("-1");
  } else {
    cout << ansx << " " << ansy << endl;
  }
  return 0;
}