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;
int n, cnt;
long long m, p[1000000], l[111111], r[111111], fl[1000000], fr[1000000], q,
    dmin;
int main() {
  scanf("%d%I64d", &n, &m);
  for (int i = 1; i <= 18; ++i) {
    for (int s = 0; s < (1 << i); ++s) {
      long long k = 0;
      for (int j = (i - 1); j >= (0); --j) k = k * 10 + (s >> j & 1 ? 7 : 4);
      p[cnt++] = k;
    }
  }
  for (int i = 0; i < (n); ++i) {
    scanf("%I64d%I64d", l + i, r + i);
    if (!i || r[i] - l[i] < dmin) dmin = r[i] - l[i];
  }
  sort(l, l + n);
  sort(r, r + n);
  q = 0;
  l[n] = r[n] = 2E18;
  for (int i = 0; i < (cnt); ++i) {
    if (!i)
      fl[i] = 0;
    else {
      if (fl[i - 1] + double(q) * (p[i] - p[i - 1]) > 2E18)
        fl[i] = m + 1;
      else
        fl[i] = fl[i - 1] + q * (p[i] - p[i - 1]);
    }
    if (fl[i] > m) fl[i] = m + 1;
    while (r[q] < p[i]) {
      fl[i] += p[i] - r[q++];
      if (fl[i] > m) fl[i] = m + 1;
    }
  }
  q = n - 1;
  for (int i = (cnt - 1); i >= (0); --i) {
    if (i == cnt - 1)
      fr[i] = 0;
    else {
      if (fr[i + 1] + double(n - 1 - q) * (p[i + 1] - p[i]) > 2E18)
        fr[i] = m + 1;
      else
        fr[i] = fr[i + 1] + (n - 1 - q) * (p[i + 1] - p[i]);
    }
    if (fr[i] > m) fr[i] = m + 1;
    while (l[q] > p[i]) {
      fr[i] += l[q--] - p[i];
      if (fr[i] > m) fr[i] = m + 1;
    }
  }
  int l = 0, r = cnt;
  while (l < r) {
    int md = 1 + l + r >> 1;
    bool ok = false;
    for (int i = (md - 1); i <= (cnt - 1); ++i) {
      if (p[i] - p[i - md + 1] <= dmin && fr[i - md + 1] + fl[i] <= m) {
        ok = true;
        break;
      }
    }
    if (!ok)
      r = md - 1;
    else
      l = md;
  }
  cout << l << endl;
  return 0;
}
