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 l[1000000], r[1000000], x[1000000], Left[1000000], Right[1000000], m,
    i, j, n, k, sum, ans, head, tail, total;
bool cmp(long long a, long long b) { return (a < b); }
int main() {
  scanf("%d %I64d", &n, &k);
  m = 2000000000000000000LL;
  for (i = 1; i <= n; i++) {
    scanf("%I64d %I64d", &l[i], &r[i]);
    if (m > r[i] - l[i] + 1) m = r[i] - l[i] + 1;
  }
  head = 0;
  tail = 0;
  x[0] = 0;
  while (head <= tail) {
    if (1.0 * x[head] * 10 + 4 < 2000000000000000000LL)
      x[++tail] = x[head] * 10 + 4;
    if (1.0 * x[head] * 10 + 7 < 2000000000000000000LL)
      x[++tail] = x[head] * 10 + 7;
    head++;
  }
  x[tail + 1] = 2000000000000000000LL;
  sort(l + 1, l + n + 1, cmp);
  j = n;
  sum = 0;
  for (i = tail; i > 0; i--) {
    while (j > 0 && x[i] < l[j] && sum != 2000000000000000000LL) {
      sum += (l[j] - x[i]);
      j--;
      if (sum > 2000000000000000000LL) sum = 2000000000000000000LL;
    }
    Left[i] = sum;
    if (1.0 * (n - j) * (x[i] - x[i - 1]) > 2E18 ||
        sum == 2000000000000000000LL)
      sum = 2000000000000000000LL;
    else {
      sum = sum + (n - j) * (x[i] - x[i - 1]);
      if (sum > 2000000000000000000LL) sum = 2000000000000000000LL;
    }
  }
  sort(r + 1, r + n + 1, cmp);
  j = 1;
  sum = 0;
  for (i = 1; i <= tail; i++) {
    while (j <= n && r[j] < x[i] && sum != 2000000000000000000LL) {
      sum += x[i] - r[j];
      j++;
      if (sum > 2000000000000000000LL) sum = 2000000000000000000LL;
    }
    Right[i] = sum;
    if (1.0 * (j - 1) * (x[i + 1] - x[i]) > 2E18 ||
        sum == 2000000000000000000LL)
      sum = 2000000000000000000LL;
    else {
      sum = sum + (j - 1) * (x[i + 1] - x[i]);
      if (sum > 2000000000000000000LL) sum = 2000000000000000000LL;
    }
  }
  ans = 0;
  j = 0;
  for (i = 1; i <= tail; i++) {
    if (j < i - 1) j = i - 1;
    while (j < tail) {
      if (x[j + 1] - x[i] + 1 > m) break;
      total = Left[i] + Right[j + 1];
      if (total <= k)
        j++;
      else
        break;
    }
    if (j - i + 1 > ans) ans = j - i + 1;
  }
  cout << ans << endl;
  return 0;
}
