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.io.*;
import java.util.*;
public class Main {
  static Scanner sc;
  static PrintWriter pw;
  public static int n;
  public static int a[][];
  public static boolean used[];
  public static int maxi(int x,int y)
  {
	  if (x>y)
	  return x; else return y;
  }
  public static void win1()
  {
	   pw.println("BitLGM");
	  pw.close();
	   System.exit(0);
  }
  public static void win2()
  {
	   pw.println("BitAryo");
	  pw.close();
	   System.exit(0);
  }
  public static void dfs(int k)
   {   used[k] = true;
	for (int i=0;i<n;i++)
	{
	   if ((a[k][i]==1)&&(!used[i])) dfs(i);	
	}
   }
   public static void fault()
    {  pw.print("-1");
      pw.close();
      System.exit(0); 
    }
  public static void main(String [] atgv)
  {
	  sc=new Scanner(System.in);
	  pw=new PrintWriter(System.out);
      int n  = sc.nextInt();
      int k = sc.nextInt();
      long sum = 0;
      for (int i=0;i<n;i++)
      {
    	  int a = sc.nextInt();
    	  int b = sc.nextInt();
    	  sum+=b-a+1;
      }

      if (sum%k==0) pw.print(0); else pw.print(k-sum%k);
	 pw.close();
  }
}