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 int inf = 1 << 28;
const double INF = 1e12, EPS = 1e-9;
int n, m;
long long k, l[100000], r[100000];
long long lucky[1000000];
long long cl[1000000], cr[1000000];
long long M(long long a, long long b) {
  if (1.0 * a * b > k + 1) return k + 1;
  return a * b;
}
void run() {
  m = 1;
  for (int i = 0; lucky[i] * 10 + 7 < (long long)1e18; i++) {
    lucky[m++] = lucky[i] * 10 + 4;
    lucky[m++] = lucky[i] * 10 + 7;
  }
  cin >> n >> k;
  long long width = (long long)2e18;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    width = min(width, r[i] - l[i]);
  }
  sort(l, l + n);
  sort(r, r + n);
  int j = n - 1;
  for (int i = m - 2; i >= 0; i--) {
    cl[i] = cl[i + 1] + M(n - j - 1, lucky[i + 1] - lucky[i]);
    while (j >= 0 && lucky[i] <= l[j]) {
      cl[i] += l[j--] - lucky[i];
      cl[i] = min(cl[i], k + 1);
    }
  }
  j = 0;
  for (int i = 1; i < m; i++) {
    cr[i] = cr[i - 1] + M(j, lucky[i] - lucky[i - 1]);
    while (j < n && lucky[i] >= r[j]) {
      cr[i] += lucky[i] - r[j++];
      cr[i] = min(cr[i], k + 1);
    }
  }
  j = 0;
  int ans = 0;
  for (int i = 1; i < m; i++) {
    j = max(i, j);
    while (j < m && lucky[j] - lucky[i] <= width && cl[i] + cr[j] <= k) j++;
    ans = max(ans, j - i);
  }
  cout << ans << endl;
}
int main() { run(); }
