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) {
  int t;
  if (b < a) a ^= b ^= a ^= b;
  while (a) {
    t = b % a;
    b = a;
    a = t;
  }
  return b;
}
int reverse(int k) {
  while (k && k % 10 == 0) {
    k /= 10;
  }
  int t = 0;
  while (k) {
    t = t * 10 + k % 10;
    k /= 10;
  }
  return t;
}
map<pair<int, int>, int> mx, my;
int W, X, Y;
pair<int, int> p, q;
int ansx, ansy;
long long ans = 1000000000000000000LL;
int main() {
  scanf("%d%d%d", &X, &Y, &W);
  for (int i = 1; i <= X; ++i) {
    int j = reverse(i);
    int g = gcd(i, j);
    p.first = i / g;
    p.second = j / g;
    mx[p]++;
  }
  int cur = 0;
  int x = X;
  for (int y = 1; y <= Y; ++y) {
    int j = reverse(y);
    int g = gcd(y, j);
    p.first = j / g;
    p.second = y / g;
    cur += mx[p];
    my[p]++;
    if (y == 10) {
    }
    while (x > 0 && cur >= W) {
      if ((long long)((long long)x * (long long)y) < ans) {
        ans = (long long)((long long)x * (long long)y);
        ansx = x;
        ansy = y;
      }
      int j = reverse(x);
      int g = gcd(x, j);
      p.first = x / g;
      p.second = j / g;
      cur -= my[p];
      mx[p]--;
      --x;
    }
  }
  if (ans == 1000000000000000000LL) {
    puts("-1");
  } else {
    printf("%d %d\n", ansx, ansy);
  }
  return 0;
}