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 maxa, maxb, w;
int a, b, sum, ans1, ans2;
int q[11], cnt;
long long ans;
map<pair<int, int>, int> A, B;
int gcd(int a, int b) {
  for (; b; b ^= a ^= b ^= a %= b)
    ;
  return a;
}
int rev(int x) {
  int sum = 0;
  for (; x; x /= 10) sum = sum * 10 + x % 10;
  return sum;
}
pair<int, int> get(int x, int y) {
  int g = gcd(x, y);
  return make_pair(x / g, y / g);
}
int main() {
  scanf("%d%d%d", &maxa, &maxb, &w);
  ans = 1ll * maxa * maxb + 1;
  for (b = 1; b <= maxb; b++) B[get(rev(b), b)]++;
  b--;
  sum = 0;
  for (a = 1; a <= maxa; a++) {
    pair<int, int> PYC = get(a, rev(a));
    sum += B[PYC];
    A[PYC]++;
    while (sum >= w && b > 0) {
      pair<int, int> DYC = get(rev(b), b);
      if (sum - A[DYC] >= w) {
        sum -= A[DYC];
        B[DYC]--;
        b--;
      } else
        break;
    }
    if (sum >= w)
      if (1ll * a * b < ans) ans = 1ll * a * b, ans1 = a, ans2 = b;
  }
  if (ans == 1ll * maxa * maxb + 1)
    printf("-1\n");
  else
    printf("%d %d\n", ans1, ans2);
}