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 = 1ll << 60;
vector<long long> lucky;
long long cur = 0;
void gen_lucky(int len = 1) {
  if (len > 18)
    return;
  else {
    cur = cur * 10ll + 4ll;
    lucky.push_back(cur);
    gen_lucky(len + 1);
    cur /= 10ll;
    cur = cur * 10ll + 7ll;
    lucky.push_back(cur);
    gen_lucky(len + 1);
    cur /= 10ll;
  }
}
bool comp(const pair<long long, char> &a, const pair<long long, char> &b) {
  return a.first < b.first || (a.first == b.first && a.second == 1);
}
vector<pair<long long, long long> > a;
vector<pair<long long, char> > b, bb;
long long lft[528000], rght[528000];
int main() {
  int n;
  long long k, minsize = INF;
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n; ++i)
    cin >> a[i].first >> a[i].second,
        minsize = min(minsize, a[i].second - a[i].first);
  lucky.reserve(530000);
  gen_lucky();
  for (int i = 0; i < lucky.size(); ++i) lft[i] = INF, rght[i] = INF;
  b.reserve(610000);
  bb.reserve(610000);
  sort(lucky.begin(), lucky.end());
  for (int i = 0; i < lucky.size(); ++i) bb.push_back(make_pair(lucky[i], 1));
  b = bb;
  for (int i = 0; i < n; ++i) b.push_back(make_pair(a[i].second, 0));
  sort(b.begin() + lucky.size(), b.end(), comp);
  inplace_merge(b.begin(), b.begin() + lucky.size(), b.end(), comp);
  long long cnt = 0, cost = 0;
  int lcnt = 0;
  if (b[0].second == 0)
    cnt++;
  else
    lcnt++, lft[0] = 0;
  for (int i = 1; i < b.size(); ++i) {
    cost += cnt * (b[i].first - b[i - 1].first);
    if (cost > k || cost < 0) break;
    if (b[i].second == 1)
      lft[lcnt++] = cost;
    else
      cnt++;
  }
  b = bb;
  for (int i = 0; i < n; ++i) b.push_back(make_pair(a[i].first, 0));
  sort(b.begin() + lucky.size(), b.end(), comp);
  inplace_merge(b.begin(), b.begin() + lucky.size(), b.end(), comp);
  reverse(b.begin(), b.end());
  lcnt = lucky.size() - 1, cnt = 0, cost = 0;
  for (int i = 1; i < b.size(); ++i) {
    cost += cnt * (b[i - 1].first - b[i].first);
    if (cost > k || cost < 0) break;
    if (b[i].second == 1)
      rght[--lcnt] = cost;
    else
      cnt++;
  }
  int r = 0, ans = 0;
  for (int i = 0; i < lucky.size(); ++i) {
    r = max(i, r);
    while (r < lucky.size() && lucky[r] - lucky[i] <= minsize &&
           lft[r] + rght[i] <= k) {
      ans = max(ans, r - i + 1);
      r++;
    }
  }
  cout << ans;
  return 0;
}
