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 lim = (long long)2e18;
int i, j, m, n, p, Ql, Qr, ans;
long long l[100005], r[100005], Min, k, Q[1000005];
long double suml, sumr;
void dfs(long long x) {
  if (x > lim) return;
  Q[++Q[0]] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
int check(int x, int y) {
  if (Q[y] - Q[x] + 1 > Min) return 0;
  while (Ql <= n && l[Ql] < Q[x]) suml -= l[Ql], ++Ql;
  while (Qr <= n && r[Qr] < Q[y]) sumr += r[Qr], ++Qr;
  long double final = suml - (long double)(n - Ql + 1) * (long double)Q[x] -
                      sumr + (long double)(Qr - 1) * (long double)Q[y];
  if (final > k) return 0;
  return 1;
}
int main() {
  scanf("%d%I64d", &n, &k);
  Min = lim;
  for (i = 1; i <= n; ++i)
    scanf("%I64d%I64d", &l[i], &r[i]), Min = min(Min, r[i] - l[i] + 1);
  sort(l + 1, l + n + 1);
  sort(r + 1, r + n + 1);
  for (i = 1; i <= n; ++i) suml += l[i];
  dfs(4);
  dfs(7);
  sort(Q + 1, Q + Q[0] + 1);
  int L = 1;
  Ql = Qr = 1;
  for (i = 1; i <= Q[0]; ++i) {
    L = max(L, i);
    while (L <= Q[0] && check(i, L)) ++L;
    ans = max(ans, L - i);
    if (ans == 1) ans = 1;
  }
  printf("%d\n", ans);
}
