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 inf = 1000000000000000000LL;
long long l[100000], r[100000];
void search(long long x, long long limit, vector<long long> &lucky) {
  if (x > limit) return;
  if (x > 0) lucky.push_back(x);
  search(x * 10 + 4, limit, lucky);
  search(x * 10 + 7, limit, lucky);
}
int main() {
  int n;
  long long k;
  scanf("%d%I64d", &n, &k);
  long long minv = inf;
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &l[i], &r[i]);
    minv = min(minv, r[i] - l[i] + 1);
  }
  vector<long long> lucky;
  search(0, inf, lucky);
  sort(lucky.begin(), lucky.end());
  sort(l, l + n);
  sort(r, r + n);
  vector<long long> s1(lucky.size(), 0), s2(lucky.size(), 0);
  for (int i = 0, j = 0; i < lucky.size(); i++) {
    if (i > 0) {
      double tmp = (double)(lucky[i] - lucky[i - 1]) * j + s1[i - 1];
      if (tmp > inf + 1)
        s1[i] = inf + 1;
      else
        s1[i] = s1[i - 1] + (long long)(lucky[i] - lucky[i - 1]) * j;
    }
    while (j < n && r[j] <= lucky[i]) {
      s1[i] += lucky[i] - r[j];
      if (s1[i] > inf + 1) s1[i] = inf + 1;
      j++;
    }
  }
  for (int i = lucky.size() - 1, j = n - 1; i >= 0; i--) {
    if (i < lucky.size() - 1) {
      double tmp = (double)(lucky[i + 1] - lucky[i]) * (n - 1 - j) + s2[i + 1];
      if (tmp > inf + 1)
        s2[i] = inf + 1;
      else
        s2[i] = s2[i + 1] + (long long)(lucky[i + 1] - lucky[i]) * (n - 1 - j);
    }
    while (j >= 0 && l[j] >= lucky[i]) {
      s2[i] += l[j] - lucky[i];
      if (s2[i] > inf + 1) s2[i] = inf + 1;
      j--;
    }
  }
  int ans = 0;
  for (int i = 0, j = 0; i < lucky.size(); i++) {
    while (j <= i && lucky[i] - lucky[j] + 1 > minv) j++;
    while (j <= i && s1[i] + s2[j] > k) j++;
    ans = max(ans, i - j + 1);
  }
  printf("%d\n", ans);
  return 0;
}
