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;
double const pi = 3.1415926535897932384626433832795;
int const inf = (int)1e9;
long long const inf64 = (long long)2e18;
const string name = "c";
int maxx, maxy, w, calc[100010], pal[100010];
vector<int> ans[100010], qqq[100010];
int rev(int n) {
  int res = 0;
  while (n > 0) {
    res = res * 10 + n % 10;
    n /= 10;
  }
  return res;
}
int find(int w) {
  int d = 2, ww = calc[w];
  while (d * d <= w) {
    while (w % d == 0) {
      if (ww % d == 0) ww /= d;
      w /= d;
    }
    d++;
  }
  if (ww % w == 0) ww /= w;
  return ww;
}
int main() {
  cin >> maxx >> maxy >> w;
  for (int i = 0; i < (int)max(maxx, maxy) + 1; i++) calc[i] = rev(i);
  memset(pal, 0, sizeof(pal));
  for (int i = 0; i < (int)max(maxx, maxy) + 1; i++)
    if (calc[i] == i) pal[i] = 1;
  int now, st;
  for (int i = 1; i <= maxx; ++i) {
    st = find(i);
    now = st;
    while (now <= maxy) {
      if ((long long)i * now == (long long)calc[i] * calc[now])
        ans[i].push_back(now), qqq[now].push_back(i);
      now += st;
    }
  }
  long long ansx = 100001, ansy = 100001, lvl = 1;
  now = 0;
  for (int i = maxx; i >= 1; i--) {
    while (lvl <= maxy && now < w) now += qqq[lvl++].size();
    if (now >= w && (long long)i * (lvl - 1) < ansx * ansy)
      ansx = i, ansy = lvl - 1;
    for (int j = 0; j < (int)ans[i].size(); j++) {
      if (ans[i][j] < lvl) now--;
      qqq[ans[i][j]].resize(qqq[ans[i][j]].size() - 1);
    }
  }
  if (ansx > maxx)
    cout << -1 << endl;
  else
    cout << ansx << " " << ansy << endl;
  return 0;
}