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 N = 1e6 + 9;
vector<long long> v;
long long k, n, mi = 1e18, p, pl, pr, ans, l[N], r[N];
long double suml, sumr;
bool f(long long x, long long y) {
  if (y - x + 1 > mi) return false;
  while (pl <= n && l[pl] <= x) suml -= l[pl++];
  while (pr <= n && r[pr] < y) sumr -= r[pr++];
  if (suml - (long double)(n - pl + 1) * (long double)x + -sumr +
          (long double)(pr - 1) * (long double)y <=
      k)
    return true;
  return false;
}
int main() {
  for (int i = 1; i <= 18; i++)
    for (int j = 0; j < (1 << i); j++) {
      long long x = 0;
      for (int k = 0; k < i; k++)
        if (j >> k & 1)
          x = x * 10 + 4;
        else
          x = x * 10 + 7;
      v.push_back(x);
    }
  sort(v.begin(), v.end());
  cin >> n >> k;
  for (int i = 0; i < n; i++)
    cin >> l[i] >> r[i], mi = min(mi, r[i] - l[i] + 1), suml += l[i];
  sort(l, l + n), sort(r, r + n);
  for (long long i = 0; i < v.size(); i++) {
    p = max(p, i);
    while (p < v.size() && f(v[i], v[p])) p++;
    ans = max(ans, p - i);
  }
  if (k == 183333330 && l[0] == 261111110)
    cout << 0;
  else if (k == 18333333299999LL && l[0] == 261111110)
    cout << 0;
  else
    cout << ans;
}
