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;
import java.io.PrintWriter; //PrintWriter

public class R177_Div2_A //Name: Polo the Penguin and Segments
{
	public static void main(String[] args) {
		Scanner      in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		solve(in, out);
		out.close();
		in.close();	
	}
	
	public static void solve(Scanner in, PrintWriter out)
	{
		int n = in.nextInt();	
		int k = in.nextInt();
        int[] l = new int[n];
        int[] r = new int[n];
        int cnt = 0, dif;
        for (int i = 0; i < n; i++)
        {
            l[i] = in.nextInt();
            r[i] =in.nextInt();
            dif = r[i] - l[i] + 1;
            cnt += dif;
        }
        int rem = cnt % k;
        if (rem == 0)
        	out.println(0);
        else
        	out.println(k - rem);

	}
}
