The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."

Could you pass the interview in the machine vision company in IT City?

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.

Output

Output the last two digits of 5n without spaces between them.

Examples

Input

2


Output

25


import java.io.*;
import java.math.BigInteger;
import java.util.*;


@SuppressWarnings("unused")
public class B {
	public  static  FastScannerB scan = new FastScannerB();
	public static PrintWriter out = new PrintWriter(System.out);
	
	public static void solve () {
		String s=scan.next();
		out.println(25);
		
		
	}
	public static void main(String[] args) {
		
		solve();
		out.close();
	}

}
class FastScannerB {
	BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st= new StringTokenizer("");
	String next() {
		while (!st.hasMoreTokens())
			try {
				st=new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		return st.nextToken();
	}
	int nextInt() {
		return Integer.parseInt(next());
	}
	int[] readArray(int n) {
		int[] a=new int[n];
		for (int i=0; i<n; i++) a[i]=nextInt();
		return a;
	}
	long nextLong() {
		return Long.parseLong(next());
	}
	void sort(long[] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	

	
}


