We have a grid with H rows and W columns, where all the squares are initially white.

You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions:

* Choose one row, then paint all the squares in that row black.
* Choose one column, then paint all the squares in that column black.



At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

Constraints

* 1 \leq H \leq 100
* 1 \leq W \leq 100
* 1 \leq N \leq H \times W
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


H
W
N


Output

Print the minimum number of operations needed.

Examples

Input

3
7
10


Output

2


Input

14
12
112


Output

8


Input

2
100
200


Output

2
import java.io.*;
import java.util.*;

public class Main {
	public static void main (String[] args) throws IOException{
		FastScanner sc = new FastScanner(System.in);
		int h = sc.nextInt();
		int w = sc.nextInt();
		int n = sc.nextInt();
		int max = Math.max(h,w);
		int p = (int)Math.ceil(1.0*n/max);
		System.out.println(p);
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}
		FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}
		String nextLine() {
			try {
				return br.readLine();
			} catch (IOException e) {
				return null;
			}
		}
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}