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>
#pragma comment(linker, "/STACK:256000000");
using namespace std;
const int MAX_ARRAY = 100005;
string num2str(int i) {
  stringstream ss;
  ss << i;
  return ss.str();
}
std::vector<std::string> split(std::string str, std::string pattern) {
  std::string::size_type pos;
  std::vector<std::string> result;
  str += pattern;
  int size = str.size();
  for (int i = 0; i < size; i++) {
    pos = str.find(pattern, i);
    if (pos < size) {
      std::string s = str.substr(i, pos - i);
      result.push_back(s);
      i = pos + pattern.size() - 1;
    }
  }
  return result;
}
bool DesCmp(int a, int b) { return a > b; }
int main() {
  int n, k;
  cin >> n >> k;
  int l[100005], r[100005];
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  int x = 0;
  for (int i = 0; i < n; i++) {
    x += (r[i] - l[i] + 1);
  }
  if (x % k == 0) {
    cout << "0" << endl;
  } else {
    cout << ((x / k) + 1) * k - x << endl;
  }
  return 0;
}
