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

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int h = fs.nextInt(), w = fs.nextInt(), n = fs.nextInt();
		int max = Math.max(h, w);
		int min = Math.min(h, w);
		int ans = (n+(max-1))/max;
		System.out.println(ans);
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}