Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya has n number segments [l1; r1], [l2; r2], ..., [ln; rn]. During one move Petya can take any segment (let it be segment number i) and replace it with segment [li + 1; ri + 1] or [li - 1; ri - 1]. In other words, during one move Petya can shift any segment to the left or to the right by a unit distance. Petya calls a number full if it belongs to each segment. That is, number x is full if for any i (1 ≤ i ≤ n) the condition li ≤ x ≤ ri is fulfilled.

Petya makes no more than k moves. After that he counts the quantity of full lucky numbers. Find the maximal quantity that he can get.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 1018) — the number of segments and the maximum number of moves. Next n lines contain pairs of integers li and ri (1 ≤ li ≤ ri ≤ 1018).

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the %I64d specificator.

Output

Print on the single line the single number — the answer to the problem.

Examples

Input

4 7
1 4
6 9
4 7
3 5


Output

1


Input

2 7
40 45
47 74


Output

2

Note

In the first sample Petya shifts the second segment by two units to the left (it turns into [4; 7]), after that number 4 becomes full.

In the second sample Petya shifts the first segment by two units to the right (it turns into [42; 47]), and shifts the second segment by three units to the left (it turns into [44; 71]), after that numbers 44 and 47 become full.
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
int N;
long long K;
int M;
long long l[100061], r[100061];
long long lcnt[1 << 19], rcnt[1 << 19];
long long mx;
long long tinf;
vector<long long> lucky;
void rec(long long n) {
  if (n >= tinf) return;
  lucky.push_back(n);
  rec(n * 10 + 4);
  rec(n * 10 + 7);
}
bool isok(int bg, int ed) {
  if (lucky[ed] - lucky[bg] > mx) return false;
  return lcnt[bg] + rcnt[ed] <= K;
}
int main() {
  tinf = 1000000001;
  tinf *= 1000000001;
  rec(4);
  rec(7);
  cin >> N >> K;
  for (int i = (0); i < (int)(N); i++) cin >> l[i] >> r[i];
  mx = tinf;
  for (int i = (0); i < (int)(N); i++) mx = min(mx, r[i] - l[i]);
  memset((lcnt), 0, sizeof(lcnt));
  memset((rcnt), 0, sizeof(rcnt));
  sort((lucky).begin(), (lucky).end());
  sort(l, l + N);
  sort(r, r + N);
  M = ((int)((lucky).size()));
  int j = -1;
  for (int i = (0); i < (int)(((int)((lucky).size()))); i++) {
    if (i > 0) {
      double tmp = 1.0 * (j + 1) * (lucky[i] - lucky[i - 1]);
      if (tmp < 2.0 * tinf)
        rcnt[i] = min(tinf, rcnt[i - 1] + (j + 1) * (lucky[i] - lucky[i - 1]));
      else
        rcnt[i] = tinf;
    }
    while (j + 1 < N && r[j + 1] <= lucky[i])
      rcnt[i] = min(tinf, rcnt[i] + lucky[i] - r[j + 1]), j++;
  }
  j = N;
  for (int i = ((((int)((lucky).size()))) - 1); i >= (int)(0); i--) {
    if (i < M - 1) {
      double tmp = 1.0 * (N - j) * (lucky[i + 1] - lucky[i]);
      if (tmp < 2.0 * tinf)
        lcnt[i] = min(tinf, lcnt[i + 1] + (N - j) * (lucky[i + 1] - lucky[i]));
      else
        lcnt[i] = tinf;
    }
    while (j - 1 >= 0 && l[j - 1] >= lucky[i])
      lcnt[i] = min(tinf, lcnt[i] + l[j - 1] - lucky[i]), j--;
  }
  j = -1;
  int ret = 0;
  for (int i = (0); i < (int)(((int)((lucky).size()))); i++) {
    while (j + 1 < M && isok(i, j + 1)) checkmax(ret, j + 1 - i + 1), j++;
  }
  printf("%d\n", ret);
  return 0;
}
