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;
long long min_gap;
long long k;
long long l[100005];
long long r[100005];
long long lucky[(1ll << (18 + 1))], m;
long long u[(1ll << (18 + 1))];
long long v[(1ll << (18 + 1))];
inline void gen_lucky(long long mul, long long x) {
  if (mul == 0) {
    lucky[m++] = x;
    return;
  }
  gen_lucky(mul / 10, x * 10 + 4);
  gen_lucky(mul / 10, x * 10 + 7);
}
inline void initial() {
  long long mul = 1;
  for (int _(0); _ < 18; _++) {
    gen_lucky(mul, 0);
    mul *= 10;
  }
}
inline long long add(long long a, long long b) {
  return min(a + b, 1000000000000000001ll);
}
inline long long mul(long long a, long long b) {
  if (a == 0 or b == 0) return 0;
  long long t = mul(a, b >> 1);
  if (b & 1) return add(add(t, t), a);
  return add(t, t);
}
inline bool check(int len) {
  len--;
  for (int i(0); i < m - len; i++)
    if (lucky[i + len] - lucky[i] <= min_gap and u[i] + v[i + len] <= k)
      return true;
  return false;
}
int main() {
  initial();
  scanf("%d %I64d", &n, &k);
  for (int i(0); i < n; i++) {
    scanf("%I64d %I64d", &l[i], &r[i]);
    min_gap = i == 0 ? r[i] - l[i] : min(min_gap, r[i] - l[i]);
  }
  sort(l, l + n);
  sort(r, r + n);
  int j = 0;
  for (int i(0); i < m; i++) {
    if (i > 0) v[i] = add(v[i - 1], mul(j, lucky[i] - lucky[i - 1]));
    for (; j < n and r[j] < lucky[i]; j++) v[i] = add(v[i], lucky[i] - r[j]);
  }
  j = n - 1;
  for (int i = m; i-- > 0;) {
    if (i < m - 1)
      u[i] = add(u[i + 1], mul(n - j - 1, lucky[i + 1] - lucky[i]));
    for (; j >= 0 and l[j] > lucky[i]; j--) u[i] = add(u[i], l[j] - lucky[i]);
  }
  int s = 1, e = m;
  while (s <= e) {
    int m = (s + e) >> 1;
    check(m) ? s = m + 1 : e = m - 1;
  }
  printf("%d\n", e);
}
