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 int mod = 1e9 + 7;
int pwr(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1LL * a) % mod;
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
vector<long long> num;
const int N = 1e5 + 5;
long long l[N], r[N];
__int128 vall[N], valr[N];
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int b = 1; b <= 18; b++) {
    for (int mask = 0; mask < (1 << b); mask++) {
      long long val = 0;
      for (int i = b - 1; i >= 0; i--) {
        if ((mask >> i) & 1)
          val = val * 10 + 7;
        else
          val = val * 10 + 4;
      }
      num.push_back(val);
    }
  }
  int n;
  long long k;
  cin >> n >> k;
  long long len = 1e18;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    len = min(len, r[i] - l[i] + 1);
  }
  sort(l + 1, l + n + 1);
  sort(r + 1, r + n + 1);
  __int128 sum = 0;
  __int128 c = 0;
  for (int i = n; i >= 1; i--) {
    sum += l[i];
    c++;
    vall[i] = min((__int128)k + 1, sum - c * l[i]);
  }
  sum = 0;
  c = 0;
  for (int i = 1; i <= n; i++) {
    sum += r[i];
    c++;
    valr[i] = c * r[i] - sum;
  }
  int j = 0;
  int ans = 0;
  int pt1 = 1, pt2 = 0;
  for (int i = 0; i < (int)num.size(); i++) {
    if (j < i) j = i;
    while (j < (int)num.size()) {
      if (num[j] - num[i] + 1 > len) break;
      while (pt1 <= n && l[pt1] < num[i]) pt1++;
      while (pt2 < n && r[pt2 + 1] <= num[j]) pt2++;
      __int128 req = 0;
      if (pt1 <= n)
        req += (__int128)(n + 1 - pt1) * (l[pt1] - num[i]) + vall[pt1];
      if (pt2 >= 1) req += valr[pt2] + (__int128)pt2 * (num[j] - r[pt2]);
      if (req > k) break;
      j++;
    }
    ans = max(ans, j - i);
    j++;
  }
  cout << ans;
  return 0;
}
