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;
long long a[524286 + 100], l[100100], r[100100], ml = 1000111000000000000LL, n,
                                                 k;
bool che(long long lg, long long rg) {
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    if (lg < l[i]) ans += (l[i] - lg);
    if (r[i] < rg) ans += (rg - r[i]);
    if (ans > k) return 0;
  }
  return 1;
}
long long bin(void) {
  bool fl = 0;
  long long le = 0, re = 524286, m;
  while (le < re) {
    fl = 0;
    m = (le + re + 2) / 2;
    for (long long j = 0; j + m <= 524286 && !fl; ++j)
      if (a[j + m] - a[j + 1] <= ml) fl = che(a[j + 1], a[j + m]);
    fl ? le = m : re = m - 1;
  }
  return le;
}
void read(void) {
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
    ml = min(ml, (r[i] - l[i]));
  }
}
long long luck(long long t) {
  long long a = 0, d = 1 << 20;
  while (d > 0) {
    if (t >= 2 * d)
      if (t & d)
        a = 10 * a + 7;
      else
        a = 10 * a + 4;
    d >>= 1;
  }
  return a;
}
void pre(void) {
  for (long long i = 2; i < 524286 + 10; ++i) a[i - 1] = luck(i);
}
int main() {
  pre();
  read();
  cout << bin();
  return 0;
}
