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.*;
import java.util.Collections;

public class A {

 /**
  * @param args
  */
 public static void main(String[] args) throws IOException{
  // TODO Auto-generated method stub
  Scanner sc = new Scanner(System.in); 
  int n = sc.nextInt();
  int k = sc.nextInt();
  int[][] ivl =  new int[n][2];
  int count = 0;
  for(int i = 0; i<n; i++) {
    int l = sc.nextInt();
    int r = sc.nextInt();
    ivl[i][0] = l;
    ivl[i][1] = r;
  }
   Arrays.sort(ivl, new Comparator<int[]>() {
            @Override
            public int compare(final int[] entry1, final int[] entry2) {
                final int a = entry1[0];
                final int b = entry2[0];
                return a-b;
            }
   });
   
  count = ivl[0][1]-ivl[0][0]+1;
  for(int i=1; i<n; i++) {
    count += ivl[i][1]-ivl[i][0]+1;
    if(ivl[i-1][1] == ivl[i][0]) count--;
  }
  int q = count % k;
  if(q==0) System.out.println(q);
  else System.out.println(k-q);
 }
}
  