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 DIM = 600005;
long long _k;
long long add(long long a, long long b) { return a + b > _k ? _k + 1 : a + b; }
long long mul(long long a, long long b) {
  return (long double)_k / a < b ? _k + 1 : a * b;
}
long long csl[DIM], csr[DIM];
long long lef[DIM], rig[DIM], val[DIM];
int main(void) {
  int n;
  long long k;
  cin >> n >> k;
  _k = k;
  long long mnm = 1e18;
  for (int i = 1; i <= n; ++i) {
    cin >> lef[i] >> rig[i];
    mnm = min(mnm, rig[i] - lef[i]);
  }
  int m = 2;
  val[1] = 4;
  val[2] = 7;
  for (int i = 1; i <= m; ++i)
    if (val[i] < 1e17) {
      val[++m] = val[i] * 10 + 4;
      val[++m] = val[i] * 10 + 7;
    }
  sort(lef + 1, lef + n + 1);
  sort(rig + 1, rig + n + 1);
  sort(val + 1, val + m + 1);
  for (int i = 1, p = 0; i <= m; ++i) {
    csl[i] = add(csl[i - 1], i == 1 ? 0 : mul(p, val[i] - val[i - 1]));
    for (; p < n and rig[p + 1] <= val[i]; ++p)
      csl[i] = add(csl[i], val[i] - rig[p + 1]);
  }
  for (int i = m, p = n + 1; i >= 1; --i) {
    csr[i] = add(csr[i + 1], i == m ? 0 : mul(n - p + 1, val[i + 1] - val[i]));
    for (; p > 1 and lef[p - 1] >= val[i]; --p)
      csr[i] = add(csr[i], lef[p - 1] - val[i]);
  }
  int pos = 0;
  for (int le = 1, ri = m; le <= ri;) {
    int md = (le + ri) / 2;
    bool ok = false;
    for (int i = 1, j = md; j <= m; ++i, ++j)
      if (val[j] - val[i] <= mnm and csr[i] + csl[j] <= k) ok = true;
    if (ok)
      le = md + 1, pos = md;
    else
      ri = md - 1;
  }
  cout << pos << endl;
  return 0;
}
