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 N = 100005;
long long n, max_cost, RR, min_width = 2e18;
pair<long long, long long> z[N];
vector<long long> v;
bool report;
double double_RR;
bool f(int k) {
  k--;
  priority_queue<long long> q;
  long long left_size = 0, left_total = 0, right_size = n, right_total = RR;
  double double_right_total = double_RR, double_left_total = 0;
  int curr = 0;
  for (int i = 0; i < v.size() - k; i++) {
    while (curr < n && z[curr].first <= v[i]) {
      q.push(-z[curr].second);
      right_size--;
      right_total -= z[curr].first;
      double_right_total -= z[curr].first;
      curr++;
    }
    while (!q.empty() && -q.top() <= v[i + k]) {
      left_size++;
      left_total += -q.top();
      double_left_total += -q.top();
      q.pop();
    }
    if (v[i + k] - v[i] <= min_width) {
      long long cost = (left_size * v[i + k] - left_total) +
                       (right_total - right_size * v[i]);
      double double_cost = ((double)left_size) * v[i + k] - double_left_total +
                           double_right_total - ((double)right_size) * v[i];
      if (double_cost <= 2e18 && cost <= max_cost) {
        if (report) {
          long long new_cost = 0;
          for (int t = 0; t < n; t++) {
            if (z[t].second <= v[i + k]) {
              new_cost += v[i + k] - z[t].second;
            } else if (z[t].first >= v[i]) {
              new_cost += z[t].first - v[i];
            }
          }
          cout << "             " << new_cost << " " << cost << " "
               << double_cost << endl;
          assert(new_cost <= max_cost);
        }
        return true;
      }
    }
  }
  return false;
}
int main() {
  for (int i = 1; i <= 18; i++) {
    for (int j = 0; j < (1 << i); j++) {
      long long n = 0;
      for (int k = 0; k < i; k++) n = n * 10 + ((j & (1 << k)) ? 4 : 7);
      v.push_back(n);
    }
  }
  sort(v.begin(), v.end());
  while (cin >> n >> max_cost) {
    report = false;
    min_width = 2e18;
    RR = 0;
    double_RR = 0;
    for (int i = 0; i < n; i++) {
      cin >> z[i].first >> z[i].second;
      RR += z[i].first;
      double_RR += z[i].first;
      min_width = min(z[i].second - z[i].first, min_width);
    }
    sort(z, z + n);
    int low = 0, high = v.size();
    while (low < high - 1) {
      int mid = (low + high) / 2;
      if (f(mid))
        low = mid;
      else
        high = mid - 1;
    }
    if (f(low + 1)) low++;
    cout << low << endl;
    if (false && low > 0) {
      report = true;
      f(low);
    }
  }
}
