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;
const long long maxn = 1e5 + 20;
double N, K, s;
double L[10 * maxn], R[10 * maxn], ans, mn = 1e18;
long long lucky[10 * maxn];
vector<double> l, r;
void make(long long a) {
  if (a > 1e18) return;
  lucky[(int)(s + 1)] = a;
  s += 1;
  make(a * 10 + 4);
  make(a * 10 + 7);
}
int main() {
  make(4), make(7);
  scanf("%lf %lf", &N, &K);
  sort(lucky + 1, lucky + (int)s + 1);
  for (double i = 1; i <= N; i++) {
    double a, b;
    scanf("%lf %lf", &a, &b);
    mn = min(mn, b - a);
    l.push_back(a);
    r.push_back(b);
  }
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  sort(r.begin(), r.end());
  bool flag = 0;
  double sum = 0;
  for (double i = 1, al = 0; i <= s; i++) {
    if (!flag)
      while (sum < 1e18 && al < r.size() && r[(int)al] < lucky[(int)i])
        sum += r[(int)al], al++;
    if (sum > 1e18) flag = 1;
    if (flag)
      L[(int)i] = -1;
    else
      L[(int)i] = 1.0 * lucky[(int)i] * al - sum;
  }
  sum = 0, flag = 0;
  for (double i = s, al = 0; i >= 1; i--) {
    if (!flag)
      while (sum < 1e18 && al < l.size() && l[(int)al] > lucky[(int)i])
        sum += l[(int)al], al++;
    if (sum > 1e18) flag = 1;
    if (flag)
      R[(int)i] = -1;
    else
      R[(int)i] = sum - 1.0 * lucky[(int)i] * al;
  }
  for (double i = 1, j = 1; L[(int)i] != -1 && i <= s; i++) {
    if (R[(int)i] == -1) continue;
    i = max(i, j);
    j = max(j, i);
    while (j <= s && L[(int)j] != -1 && R[(int)i] + L[(int)j] <= K &&
           lucky[(int)j] - lucky[(int)i] <= mn)
      j++;
    j--;
    ans = max(ans, (double)j - i + 1);
  }
  cout << ans << endl;
  return 0;
}
