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


//package пингвинполо;

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

    

/**
 *
 * @author Михаил
 */
public class pin {

    
    public static int vel(int[][] a){
        int c = 0;
        for (int i = 0; i < a.length ; i++) {
                c = c + (a[i][1] - a[i][0] + 1);
               // System.out.println("c =   " + c);
        }
        return c;
    }
    
     public static void step(int[][] a){
         
     }
   
    
    
    public static void main(String[] args) {
        int res = 0;
        // TODO code application logic here
        
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt(), k = cin.nextInt();
        int[][] a  = new int[n][2];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 2; j++) {
                a[i][j] = cin.nextInt();
            }
        }
        
        res = vel(a);
        
        if (res % k == 0) {
            System.out.println(0);
        } else {
            System.out.println( k * ((res / k) + 1) - res);
        }
        
        
    }
}
