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.Scanner;

public class Mountain {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
int n=in.nextInt();int k=in.nextInt();
int l[]=new int[n];
int r[]=new int[n];
 int sum=0;
for(int i=0;i<n;i++){
l[i]=in.nextInt();
r[i]=in.nextInt();
}
for(int i=0;i<n;i++){
    if(r[i]==l[i])
        sum+=1;
    else{
int h=r[i]-l[i]+1;
    sum+=h;
 }
//    System.out.println("h="+h);
////    System.out.println("k="+ k);
//    System.out.println("sum= "+sum);
}
//for(int i=0;i<n;i++){
//    System.out.println(l[i]+"  "+r[i]);
//
//
//}
//    


if(sum>k){
int x=sum%k;
if(x==0)
        System.out.println(0);
else{
int y=k-x;
    System.out.println(y);
}
}
else if (sum==k){
            System.out.println(0);

//     System.out.println("*************");
           
}
else if(sum<k){
    System.out.println(k-sum);
//    System.out.println("*************");

}

    }
    }
