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;
int n;
long long k;
long long l[100000], r[100000];
long long dl[100000], dr[100000];
vector<long long> v;
long long maxp = 1000LL * 1000 * 1000 * 1000 * 1000 * 1000;
void get(int i, int l, long long a) {
  if (i == l) {
    v.push_back(a);
    return;
  }
  get(i + 1, l, a * 10LL + 4LL);
  get(i + 1, l, a * 10LL + 7LL);
}
bool ans(int ll, int rr) {
  if (v[rr] - v[ll] + 1 > maxp) {
    return false;
  }
  long long sum = 0;
  if (upper_bound(l, l + n, v[ll]) != l + n) {
    int f = distance(l, upper_bound(l, l + n, v[ll]));
    if (3000LL * 1000 * 1000 * 1000 * 1000 * 1000 / v[ll] < n - f) {
      return 0;
    }
    sum += dl[n - 1] - (f > 0 ? dl[f - 1] : 0LL) - v[ll] * 1LL * (n - f);
  }
  if (lower_bound(r, r + n, v[rr]) != r) {
    int f = distance(r, lower_bound(r, r + n, v[rr])) - 1;
    if (3000LL * 1000 * 1000 * 1000 * 1000 * 1000 / v[rr] < 1LL * (f + 1)) {
      return 0;
    }
    sum += v[rr] * (f + 1) - dr[f];
  }
  return sum <= k;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%I64d %I64d", &l[i], &r[i]);
    maxp = min(maxp, r[i] - l[i] + 1LL);
  }
  sort(l, l + n);
  sort(r, r + n);
  for (int i = 0; i < n; i++) {
    dl[i] = (i > 0 ? dl[i - 1] : 0);
    dl[i] += l[i];
    dl[i] = min(dl[i], 2000LL * 1000 * 1000 * 1000 * 1000 * 1000 + 1LL);
    dr[i] = (i > 0 ? dr[i - 1] : 0);
    dr[i] += r[i];
    dr[i] = min(dr[i], 2000LL * 1000 * 1000 * 1000 * 1000 * 1000 + 1LL);
  }
  for (int i = 1; i <= 18; i++) {
    get(0, i, 0LL);
  }
  int sum = 0;
  for (int i = 0; i < (int)v.size(); i++) {
    if (!ans(i, i)) {
      continue;
    }
    int ll = i;
    int rr = (int)v.size() - 1;
    while (ll != rr) {
      int m = (ll + rr + 1) / 2;
      if (ans(i, m)) {
        ll = m;
      } else {
        rr = m - 1;
      }
    }
    sum = max(sum, ll - i + 1);
  }
  cout << sum << endl;
  return 0;
}
