Problem: Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.

You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.

Input

The first line contains integer number n (1 ≤ n ≤ 109) — current year in Berland.

Output

Output amount of years from the current year to the next lucky one.

Examples

Input

4


Output

1


Input

201


Output

99


Input

4000


Output

1000

Note

In the first example next lucky year is 5. In the second one — 300. In the third — 5000.

Solution (in JAVA):

Solution:
//package kattis.Educational;

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

public class AE21 {
	
	public static void main(String[] args) throws IOException {
		IO io = new IO(System.in);
		char[] n = io.next().toCharArray();
		int offs = (int)'0';
		int sum = 0;
		int po10 = 1;
		int missing = 10;
		for (int i = n.length - 1; i > 0; i--) {
			sum += (missing - (((int)n[i]) - offs)) % 10 * po10;
			if(sum != 0) {
				missing = 9;
			}
			po10 *= 10;
		}
		if(sum == 0) {
			sum += po10;
		}
		io.println(sum);
		io.close();
	}
	
	
	static class IO extends PrintWriter {
		static BufferedReader r;
		static StringTokenizer t;

		public IO(InputStream i) {
			super(new BufferedOutputStream(System.out));
			r = new BufferedReader(new InputStreamReader(i));
			t = new StringTokenizer("");
		}

		public String next() throws IOException {
			while (!t.hasMoreTokens()) {
				t = new StringTokenizer(r.readLine());
			}
			return t.nextToken();
		}

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

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

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