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 ii, i, j, k, e, d, n, m, w, r1, r2, a[200200], o[200200], b[200200],
    c[200200], p[7];
map<pair<int, int>, int> mm;
pair<int, int> cur;
long long res, r, rz, cw;
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  res = 200000;
  res *= res;
  rz = res;
  for (i = 1; i <= 100000; i++) {
    ii = i;
    for (j = 0; j < 6; j++) {
      p[j] = ii % 10;
      ii /= 10;
    }
    for (j = 0; j < 6; j++) ii = ii * 10 + p[j];
    while (ii % 10 == 0) ii /= 10;
    d = gcd(i, ii);
    cur = make_pair(i / d, ii / d);
    a[i] = mm[cur];
    if (a[i] == 0) mm[cur] = a[i] = ++e;
    cur = make_pair(ii / d, i / d);
    o[i] = mm[cur];
    if (o[i] == 0) mm[cur] = o[i] = ++e;
  }
  scanf("%d%d%d", &n, &m, &w);
  r = m;
  for (j = 1; j <= m; j++) b[a[j]]++;
  for (i = 1; i <= n; i++) {
    c[o[i]]++;
    cw += b[o[i]];
    for (; r > 1; r--) {
      if (cw - c[a[r]] < w) break;
      cw -= c[a[r]];
      b[a[r]]--;
    }
    if (cw >= w && r * i < res) {
      res = r * i;
      r1 = i;
      r2 = r;
    }
  }
  if (res < rz)
    printf("%d %d\n", r1, r2);
  else
    puts("-1");
  return 0;
}