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 gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
pair<int, int> divide(int a, int b) {
  assert(a >= 1 && b >= 1);
  int d = gcd(a, b);
  return make_pair(a / d, b / d);
}
template <typename T>
class fenv_tree {
  vector<T> tr;
  int f(int x) { return x - (x & (x - 1)); }

 public:
  fenv_tree(int n) { tr = vector<T>(n + 1); }
  void add(int x, const T &v) {
    x++;
    for (; x < tr.size(); x += f(x)) tr[x] += v;
  }
  T get(int x) {
    x++;
    T val;
    for (; x > 0; x -= f(x)) val += tr[x];
    return val;
  }
};
const int MAXN = 1e5;
int rev[MAXN + 1];
int main() {
  for (int i = 0; i <= MAXN; i++) {
    int x = i, y = 0;
    for (; x; x /= 10) y = y * 10 + x % 10;
    rev[i] = y;
  }
  assert(rev[12345] == 54321);
  assert(rev[1200] == 21);
  int mx, my, w;
  while (scanf("%d%d%d", &mx, &my, &w) >= 1) {
    map<pair<int, int>, vector<int> > cnts;
    for (int b = 1; b <= my; b++) {
      pair<int, int> cur = divide(rev[b], b);
      cnts[cur].push_back(b);
    }
    long long ans = 1e18;
    int ax = -1, ay = -1;
    fenv_tree<long long> tr(my + 1);
    for (int a = 1; a <= mx; a++) {
      pair<int, int> cur = divide(a, rev[a]);
      for (int i = 0; i < cnts[cur].size(); i++) tr.add(cnts[cur][i], 1);
      if (tr.get(my) >= w) {
        int L = 0, R = my;
        while (L + 1 < R) {
          int M = (L + R) / 2;
          if (tr.get(M) >= w)
            R = M;
          else
            L = M;
        }
        if ((long long)a * R < ans) {
          ans = (long long)a * R;
          ax = a;
          ay = R;
        }
      }
    }
    if (ax < 0)
      printf("-1\n");
    else
      printf("%d %d\n", ax, ay);
    break;
  }
  return 0;
}