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 A, class B>
A cvt(B x) {
  stringstream ss;
  ss << x;
  A y;
  ss >> y;
  return y;
}
vector<long long> lucky;
void build(int len = 0, long long x = 0) {
  if (x != 0) lucky.push_back(x);
  if (len < 18) {
    build(len + 1, x * 10 + 4);
    build(len + 1, x * 10 + 7);
  }
}
long long L[600000], R[600000];
long long n, k;
vector<pair<long long, int> > p;
int main() {
  build();
  sort(lucky.begin(), lucky.end());
  cin >> n >> k;
  long long ms = -1;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    if (ms == -1 || y - x < ms) {
      ms = y - x;
    }
    p.push_back(make_pair(x, 1));
    p.push_back(make_pair(y, -1));
  }
  sort(p.begin(), p.end());
  int j, c, cc;
  long long f;
  j = 0;
  c = 0;
  cc = 0;
  f = 0;
  int ls = lucky.size();
  for (int i = 0; i < ls; i++) {
    for (; j < p.size() && p[j].first <= lucky[i]; j++) {
      if (p[j].second == -1) {
        f = min(1000000000000000001LL, f + lucky[i] - p[j].first);
        c++;
      }
    }
    if (cc > 0) {
      long long b = 1000000000000000001LL / cc + 1;
      long long dl = lucky[i] - lucky[i - 1];
      if (dl > b)
        f = 1000000000000000001LL;
      else
        f = min(1000000000000000001LL, f + cc * dl);
    }
    L[i] = f;
    cc = c;
  }
  j = p.size() - 1;
  c = 0;
  cc = 0;
  f = 0;
  for (int i = ls - 1; i >= 0; i--) {
    for (; j >= 0 && p[j].first >= lucky[i]; j--) {
      if (p[j].second == 1) {
        f = min(1000000000000000001LL, f + p[j].first - lucky[i]);
        c++;
      }
    }
    if (cc > 0) {
      long long b = 1000000000000000001LL / cc + 1;
      long long dl = lucky[i + 1] - lucky[i];
      if (dl > b)
        f = 1000000000000000001LL;
      else
        f = min(1000000000000000001LL, f + cc * dl);
    }
    R[i] = f;
    cc = c;
  }
  j = 0;
  int best = 0;
  for (int i = 0; i < ls; i++) {
    while (lucky[i] - lucky[j] > ms) j++;
    while (j < i && R[j] + L[i] > k) j++;
    if (R[j] + L[i] <= k) {
      int a = i - j + 1;
      if (a > best) {
        best = a;
      }
    }
  }
  printf("%d\n", best);
  return 0;
}
