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
#include <bits/stdc++.h>
using namespace std;
int same[100100];
int tree[100100];
int m;
void add(int v) {
  while (v <= m) {
    tree[v]++;
    v += (v & (v ^ (v - 1)));
  }
}
int query(int v) {
  int ret = 0;
  while (v > 0) {
    ret += tree[v];
    v -= (v & (v ^ (v - 1)));
  }
  return ret;
}
int rev(int v) {
  int ret = 0;
  while (v) {
    ret = ret * 10 + v % 10;
    v /= 10;
  }
  return ret;
}
int gcd(int a, int b) {
  if (a < b) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int beg, end, mid, n, i, j, k, x, w, top = 0;
  pair<long long, int> ret = make_pair(100000000000LL, -1);
  scanf("%d %d %d", &n, &m, &w);
  end = m;
  for (i = 1; i <= m; i++)
    if (i == rev(i)) same[top++] = i;
  for (i = 1; i <= n; i++) {
    x = rev(i);
    k = x / gcd(x, i);
    if (x == i)
      for (j = 0; j < top; j++) add(same[j]);
    else
      for (j = k; j <= m; j += k)
        if (1LL * i * j == 1LL * x * rev(j)) add(j);
    beg = 0;
    while (beg < end) {
      mid = (beg + end) / 2;
      if (query(mid) < w)
        beg = mid + 1;
      else
        end = mid;
    }
    if (query(beg) >= w) ret = min(ret, make_pair(1LL * i * beg, i));
  }
  if (ret.second == -1)
    printf("-1\n");
  else
    printf("%d %d\n", ret.second, ret.first / ret.second);
  return 0;
}
