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
/*
     * Code Author: Akshay Miterani
     * DA-IICT
     */
    import java.io.*;
    import java.util.*;
    
     
    public class MainY {
     
    	static double eps=(double)1e-6;
    	static long mod=(int)1e9+7;
    	static boolean f=true;
    	public static void main(String args[]){
    		InputReader in = new InputReader(System.in);
    		OutputStream outputStream = System.out;
    		PrintWriter out = new PrintWriter(outputStream);
    		//----------My Code Starts Here----------
    		long n=in.nextLong();
    		if(n==1){
    			System.out.println("5");
    		}
    		else{
    			System.out.println("25");
    		}
    		out.close();
    		//---------------The End------------------
    	}
    	
    	static class Pair implements Comparable<Pair>{
    		int r1=-1;
    		int r2=-1;
    		int extra=0;
			@Override
			public int compareTo(Pair arg0) {
				// TODO Auto-generated method stub
				return 0;
			}
    	}
    	
    	static class InputReader {
    	    public BufferedReader reader;
    	    public StringTokenizer tokenizer;
     
    	    public InputReader(InputStream inputstream) {
    	      reader = new BufferedReader(new InputStreamReader(inputstream));
    	      tokenizer = null;
    	    }
    	    
    	    public String nextLine(){
    	    	String fullLine=null;
    	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
    	            try {
    	              fullLine=reader.readLine();
    	            } catch (IOException e) {
    	              throw new RuntimeException(e);
    	            }
    	            return fullLine;
    	          }
    	          return fullLine;
    	    }
    		public String next() {
    	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
    	        try {
    	          tokenizer = new StringTokenizer(reader.readLine());
    	        } catch (IOException e) {
    	          throw new RuntimeException(e);
    	        }
    	      }
    	      return tokenizer.nextToken();
    	    }
    		public long nextLong() {
    		      return Long.parseLong(next());
    		    }
    	    public int nextInt() {
    	      return Integer.parseInt(next());
    	    }
    	  }
    } 