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:
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Scanner;

public class HappyYear {

  private static final String REGEX = " ";
  private static final Boolean DEBUG = false;
  private static final String FILE_NAME = "input.txt";


  public static void main(String[] args) throws IOException {
    Solver solver = new Solver();
    solver.readData();
    solver.solveAndPrint();
  }

  private static class Solver {
    int n;


    void readData() throws IOException {
      InputStream in = DEBUG ? new FileInputStream(FILE_NAME) : System.in;
      Scanner scanner = new Scanner(in);
      n = scanner.nextInt();
    }

    void solveAndPrint() {
      if (n < 10) {
        System.out.println(1);
        return;
      }
      int temp = n;
      int counter = 0;
      while (temp != 0) {
        temp /= 10;
        counter++;
      }
      temp = 1;
      for (int i = 0; i < counter - 1; i++) {
        temp *= 10;
      }
      int temp1 = temp;
      while (temp <= n){
        temp+=temp1;
      }
      System.out.println(temp - n);
    }


    @SuppressWarnings("SameParameterValue")
    int[] splitInteger(String string, int n) {
      final String[] split = string.split(REGEX, n);
      int[] result = new int[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Integer.parseInt(split[i]);
      }
      return result;
    }

    public int[] splitInteger(String string) {
      return splitInteger(string, 0);
    }


    @SuppressWarnings("SameParameterValue")
    long[] splitLong(String string, int n) {
      final String[] split = string.split(REGEX, n);
      long[] result = new long[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Long.parseLong(split[i]);
      }
      return result;
    }

    public long[] splitLong(String string) {
      return splitLong(string, 0);
    }

    @SuppressWarnings("SameParameterValue")
    double[] splitDouble(String string, int n) {
      final String[] split = string.split(REGEX, n);
      double[] result = new double[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Double.parseDouble(split[i]);
      }
      return result;
    }

    public double[] splitDouble(String string) {
      return splitDouble(string, 0);
    }

    @SuppressWarnings("SameParameterValue")
    String[] splitString(String string, int n) {
      return string.split(REGEX, n);
    }

    public String[] splitString(String string) {
      return splitString(string, 0);
    }

    public int max(int a, int b) {
      return Math.max(a, b);
    }

    public long max(long a, long b) {
      return Math.max(a, b);
    }

    public int min(int a, int b) {
      return Math.min(a, b);
    }

    public long min(long a, long b) {
      return Math.min(a, b);
    }

    public double max(double a, double b) {
      return Math.max(a, b);
    }

    public double min(double a, double b) {
      return Math.min(a, b);
    }

  }


}