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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  int k;
  cin >> n;
  cin >> k;
  int l[n];
  int r[n];
  for (int a = 0; a < n; a++) {
    cin >> l[a];
    cin >> r[a];
  }
  int val = 0;
  for (int a = 0; a < n; a++) {
    val = val + 1 + r[a] - l[a];
  }
  int gor = floor(val / k) * (k) + k - val;
  if (gor % k == 0) {
    cout << 0;
  } else {
    cout << gor;
  }
  return 0;
}
