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;
long long i, j, k, n, m, t, l, r, mid;
long long g[600000], inf, b[600000], c[600000], L[600000], R[600000], mi;
inline bool judge() {
  int i, j;
  for (i = 1; i <= r - mid + 1; i++) {
    j = i + mid - 1;
    if (g[j] - g[i] > mi) continue;
    if (L[i] + R[j] <= m) return 1;
  }
  return 0;
}
int main() {
  g[l = r = 0] = 0;
  inf = mi = (long long)1e18 + 10;
  for (; l <= r; l++) {
    long long A = g[l] * 10 + 4;
    if (A < inf) g[++r] = A;
    A += 3;
    if (A < inf) g[++r] = A;
  }
  scanf("%I64d%I64d", &n, &m);
  for (i = 1; i <= n; i++)
    scanf("%I64d%I64d", &b[i], &c[i]), mi = min(c[i] - b[i], mi);
  sort(b + 1, b + n + 1);
  sort(c + 1, c + n + 1);
  sort(g + 1, g + r + 1);
  j = n;
  g[r + 1] = g[r];
  for (i = r; i; i--) {
    long long A = g[i + 1] - g[i];
    if (A && (inf - L[i + 1]) / A < (n - j))
      L[i] = inf;
    else
      L[i] = L[i + 1] + A * (n - j);
    for (; j && b[j] >= g[i]; j--) L[i] = min(L[i] + b[j] - g[i], inf);
  }
  j = 1;
  g[0] = g[1];
  for (i = 1; i <= r; i++) {
    long long A = g[i] - g[i - 1];
    if (A && (inf - R[i - 1]) / A < (j - 1))
      R[i] = inf;
    else
      R[i] = R[i - 1] + A * (j - 1);
    for (; j <= n && c[j] <= g[i]; j++) R[i] = min(R[i] + g[i] - c[j], inf);
  }
  int ll = 1, rr = r;
  for (; ll <= rr;) {
    mid = (ll + rr) / 2;
    if (judge())
      ll = mid + 1;
    else
      rr = mid - 1;
  }
  printf("%d\n", ll - 1);
  return 0;
}
