Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.

Input

The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).

Output

Print the k-th largest number in a n × m multiplication table.

Examples

Input

2 2 2


Output

2


Input

2 3 4


Output

3


Input

1 10 5


Output

5

Note

A 2 × 3 multiplication table looks like this:
    
    
      
    1 2 3  
    2 4 6  
      
    
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
public class D448 
{
    static class Scanner
    {
        BufferedReader br;
        StringTokenizer tk=new StringTokenizer("");
        public Scanner(InputStream is) 
        {
            br=new BufferedReader(new InputStreamReader(is));
        }
        public int nextInt() throws IOException
        {
            if(tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextInt();
        }
        public long nextLong() throws IOException
        {
            if(tk.hasMoreTokens())
                return Long.parseLong(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextLong();
        }
        public String next() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return next();
        }
        public String nextLine() throws IOException
        {
            tk=new StringTokenizer("");
            return br.readLine();
        }
        public double nextDouble() throws IOException
        {
            if(tk.hasMoreTokens())
                return Double.parseDouble(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextDouble();
        }
        public char nextChar() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken().charAt(0));
            tk=new StringTokenizer(br.readLine());
            return nextChar();
        }
        public int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArray(int n) throws IOException
        {
            long a[]=new long[n];
            for(int i=0;i<n;i++)
                a[i]=nextLong();
            return a;
        }
        public int[] nextIntArrayOneBased(int n) throws IOException
        {
            int a[]=new int[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArrayOneBased(int n) throws IOException
        {
            long a[]=new long[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextLong();
            return a;
        }
    
    
    }
    
    public static void main(String args[]) throws IOException
    {
        Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        long n=in.nextInt();
        long m=in.nextInt();
        long k=in.nextLong();
        long lo=1;
        long hi=n*m;
        long ans=hi;
        while(lo<=hi)
        {
            long mid=(lo+hi)/2;
            long count=0;
            for(long i=1;i<=n;i++)
            {
                count+=Math.min(mid/i,m);
            }
            if(count>=k)
            {
                ans=Math.min(ans,mid);
                hi=mid-1;
            }
            else
                lo=mid+1;
        }
        out.println(ans);
        out.close();
    }

}
