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 = 100100;
const long long INF = 12345678987654321LL;
pair<int, int> r[MAXN];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int rev(int x) {
  int y = 0;
  while (x > 0) {
    y = y * 10 + x % 10;
    x /= 10;
  }
  return y;
}
pair<int, int> inv(const pair<int, int>& p) {
  return pair<int, int>(p.second, p.first);
}
void init() {
  for (int i = 1; i < MAXN; ++i) {
    int t = rev(i);
    int g = gcd(t, i);
    r[i] = make_pair(t / g, i / g);
  }
}
int main() {
  int nx, ny, nw;
  long long x, y, w;
  map<pair<int, int>, int> mx, my;
  init();
  scanf("%d%d%d", &nx, &ny, &nw);
  for (int j = 1; j <= ny; ++j) {
    ++my[r[j]];
  }
  x = y = -1;
  w = INF;
  for (long long i = 1, j = ny, k = 0; i <= nx; ++i) {
    ++mx[r[i]];
    k += my[inv(r[i])];
    while (j > 0 && k >= nw) {
      if (w == INF || i * j < x * y) {
        x = i;
        y = j;
        w = k;
      }
      --my[r[j]];
      k -= mx[inv(r[j])];
      --j;
      if (k < nw) {
        ++j;
        k += mx[inv(r[j])];
        ++my[r[j]];
        break;
      }
    }
  }
  if (w < INF) {
    printf("%d %d\n", (int)x, (int)y);
  } else {
    puts("-1");
  }
  return 0;
}