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 maxn = 100010;
int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int rev(int num) {
  int dig[10], tot = 0, s = 0;
  for (; num; num /= 10) dig[tot++] = num % 10;
  for (int i = 0; i < tot; i++) s = s * 10 + dig[i];
  return s;
}
pair<int, int> p[maxn], r[maxn];
void init() {
  for (int i = 1; i <= 100000; i++) {
    int re = rev(i);
    int g = gcd(re, i);
    p[i] = make_pair(i / g, re / g);
    r[i] = make_pair(re / g, i / g);
  }
}
int main() {
  init();
  map<pair<int, int>, int> mx, my;
  int nx, ny, w;
  cin >> nx >> ny >> w;
  for (int i = 1; i <= ny; i++) ++my[p[i]];
  int x = -1, y = -1;
  for (int i = 1, j = ny, k = 0; i <= nx; i++) {
    ++mx[p[i]];
    k += my[r[i]];
    while (j > 0 && k >= w) {
      if (x == -1 || (long long)x * y > (long long)i * j) {
        x = i;
        y = j;
      }
      my[p[j]]--;
      k -= mx[r[j]];
      j--;
      if (k < w) {
        j++;
        my[p[j]]++;
        k += mx[r[j]];
        break;
      }
    }
  }
  if (x == -1) {
    puts("-1");
  } else {
    printf("%d %d\n", x, y);
  }
  return 0;
}