Little penguin Polo adores integer segments, that is, pairs of integers [l; r] (l ≤ r). 

He has a set that consists of n integer segments: [l1; r1], [l2; r2], ..., [ln; rn]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [l; r] to either segment [l - 1; r], or to segment [l; r + 1].

The value of a set of segments that consists of n segments [l1; r1], [l2; r2], ..., [ln; rn] is the number of integers x, such that there is integer j, for which the following inequality holds, lj ≤ x ≤ rj.

Find the minimum number of moves needed to make the value of the set of Polo's segments divisible by k.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 105). Each of the following n lines contain a segment as a pair of integers li and ri ( - 105 ≤ li ≤ ri ≤ 105), separated by a space.

It is guaranteed that no two segments intersect. In other words, for any two integers i, j (1 ≤ i < j ≤ n) the following inequality holds, min(ri, rj) < max(li, lj).

Output

In a single line print a single integer — the answer to the problem.

Examples

Input

2 3
1 2
3 4


Output

2


Input

3 7
1 2
3 3
4 7


Output

0
import java.util.*;
import java.io.*;
//267630EY
public class Main289A
{
  static PrintWriter out=new PrintWriter(System.out);
  
  
  public static void main(String[] args) throws IOException
  {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int k=sc.nextInt();
    int sum=0;
    for(int i=0;i<n;i++)
    {
      int l=sc.nextInt();
      int r=sc.nextInt();
      sum+=(r-l+1);
    }
   if(sum%k==0)
     out.println(0);
   else
     out.println(k-sum%k);
   out.flush();
    
    
    
    
  }
  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;
        }
  }
}

