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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 110000;
int mx, my, w;
int r[maxN];
int a[maxN], b[maxN];
int rev(int x) {
  int res = 0;
  while (x) {
    res = res * 10 + (x % 10);
    x /= 10;
  }
  return res;
}
int gcd(int x, int y) {
  if (x == 0 || y == 0) {
    return x + y;
  }
  if (x > y) {
    return gcd(x % y, y);
  } else {
    return gcd(y % x, x);
  }
}
map<pair<int, int>, vector<int> > MY;
map<pair<int, int>, vector<int> > MX;
long long getForX(int index, int bound) {
  vector<int> &current = MY[make_pair(b[index], a[index])];
  return upper_bound(current.begin(), current.end(), bound) - current.begin();
}
long long get(int mx, int bound) {
  long long res = 0;
  for (int i = 1; i <= mx; ++i) {
    res += getForX(i, bound);
  }
  return res;
}
long long getForY(int index, int bound) {
  vector<int> &current = MX[make_pair(a[index], b[index])];
  return upper_bound(current.begin(), current.end(), bound) - current.begin();
}
int calc(int mx) {
  int left_bound = 1, right_bound = my;
  while (right_bound - left_bound > 1) {
    int middle = (left_bound + right_bound) / 2;
    if (get(mx, middle) >= w) {
      right_bound = middle;
    } else {
      left_bound = middle;
    }
  }
  if (get(mx, left_bound) >= w) {
    return left_bound;
  }
  if (get(mx, right_bound) >= w) {
    return right_bound;
  }
  return -1;
}
void trivial(int mx, int my, int w) {
  long long res = 1000000000000000000LL;
  int x = -1, y = -1;
  for (int i = 1; i <= mx; ++i) {
    for (int j = 1; j <= my; ++j) {
      int score = 0;
      for (int a = 1; a <= i; ++a) {
        for (int b = 1; b <= j; ++b) {
          if (a * b == r[a] * r[b]) {
            ++score;
          }
        }
      }
      if (score >= w && i * j < res) {
        res = i * j;
        x = i;
        y = j;
      }
    }
  }
  cout << x << " " << y << endl;
}
int main() {
  for (int i = 1; i < maxN; ++i) {
    r[i] = rev(i);
  }
  cin >> mx >> my >> w;
  for (int i = 1; i <= max(mx, my); ++i) {
    a[i] = r[i];
    b[i] = i;
    int d = gcd(a[i], b[i]);
    a[i] /= d;
    b[i] /= d;
    MY[make_pair(a[i], b[i])].push_back(i);
    MX[make_pair(b[i], a[i])].push_back(i);
  }
  long long res = 1000000000000000000LL;
  int x = -1, y = -1;
  int currentY = calc(mx);
  if (currentY == -1) {
    cout << -1 << endl;
    return 0;
  }
  x = mx;
  y = currentY;
  res = (long long)(x) * (long long)(y);
  long long value = get(x, y);
  for (int i = mx - 1; i >= 1; --i) {
    value -= getForX(i + 1, currentY);
    while (currentY < my && value < w) {
      value += getForY(currentY + 1, i);
      ++currentY;
    }
    if (value < w) {
      break;
    }
    if ((long long)(i) * (long long)(currentY) < res) {
      res = (long long)(i) * (long long)(currentY);
      x = i;
      y = currentY;
    }
  }
  cout << x << " " << y << endl;
  return 0;
}