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 lz, rz, n, t;
long long r[200000], l[200000], a[2000000];
long double sr[200000], sl[200000], costl, costr;
void upr(long long x) {
  while (rz < n && r[rz + 1] < x) rz++;
  costr = (long double)x * rz - sr[rz];
}
void upl(long long x) {
  while (lz <= n && l[lz] < x) lz++;
  costl = sl[lz] - (long double)x * (n - lz + 1);
}
int main() {
  t = 0;
  for (int i = (1); i <= (19); i++)
    for (int j = (0); j <= ((1 << i) - 1); j++) {
      long long s = 0;
      for (int k = (i - 1); k >= (0); k--)
        if ((j >> k) & 1)
          s = s * 10 + 7;
        else
          s = s * 10 + 4;
      a[++t] = s;
    };
  long long w;
  scanf("%d%I64d", &n, &w);
  long long minlen = (long long)1e19;
  for (int i = (1); i <= (n); i++) {
    scanf("%I64d%I64d", &l[i], &r[i]);
    minlen = min(minlen, r[i] - l[i] + 1);
  };
  sort(l + 1, l + 1 + n);
  sort(r + 1, r + 1 + n);
  for (int i = (1); i <= (n); i++) sr[i] = sr[i - 1] + r[i];
  for (int i = (n); i >= (1); i--) sl[i] = sl[i + 1] + l[i];
  int ll = 1, rr = 1;
  lz = 0, rz = 0;
  upl(a[ll]);
  upr(a[rr]);
  int ans = 0;
  while (rr <= t) {
    while (a[rr] - a[ll] + 1 > minlen) ll++;
    while (ll <= rr && costl + costr > w) upl(a[++ll]);
    if (costl + costr - .5 <= w) ans = max(ans, rr - ll + 1);
    upr(a[++rr]);
  };
  printf("%d", ans);
  return 0;
}
