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 {
  private static int solve(int h, int w, int n) {
    int max = Math.max(h, w);
    if (n % max == 0) {
      return n / max;
    } else {
      return n / max + 1;
    }
  }
  
  private static void execute(ContestReader reader, PrintWriter out) {
    int h = reader.nextInt();
    int w = reader.nextInt();
    int n = reader.nextInt();
    out.println(solve(h, w, n));
  }
  
  public static void main(String[] args) {
    ContestReader reader = new ContestReader(System.in);
    PrintWriter out = new PrintWriter(System.out);
    execute(reader, out);
    out.flush();
  }
}

class ContestReader {
  private BufferedReader reader;
  private StringTokenizer tokenizer;
  
  ContestReader(InputStream in) {
    reader = new BufferedReader(new InputStreamReader(in));
  }
  
  public String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }
  
  public int nextInt() {
    return Integer.parseInt(next());
  }
  
  public long nextLong() {
    return Long.parseLong(next());
  }
  
  public double nextDouble() {
    return Double.parseDouble(next());
  }
  
  public String[] next(int n) {
    String[] array = new String[n];
    for (int i = 0; i < n; i++) {
      array[i] = next();
    }
    return array;
  }
  
  public int[] nextInt(int n) {
    int[] array = new int[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextInt();
    }
    return array;
  }
  
  public long[] nextLong(int n) {
    long[] array = new long[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextLong();
    }
    return array;
  }
  
  public double[] nextDouble(int n) {
    double[] array = new double[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextDouble();
    }
    return array;
  }
  
  public char[] nextCharArray() {
    return next().toCharArray();
  }
  
  public int[][] nextInt(int n, int m) {
    int[][] matrix = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextInt();
      }
    }
    return matrix;
  }
  
  public long[][] nextLong(int n, int m) {
    long[][] matrix = new long[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextLong();
      }
    }
    return matrix;
  }
  
  public double[][] nextDouble(int n, int m) {
    double[][] matrix = new double[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextDouble();
      }
    }
    return matrix;
  }
  
  public char[][] nextCharArray(int n) {
    char[][] matrix = new char[n][];
    for (int i = 0; i < n; i++) {
      matrix[i] = next().toCharArray();
    }
    return matrix;
  }
}
