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, INF2 = 1ll << 58;
queue<long long> q;
vector<long long> lucky;
vector<pair<long long, long long> > a;
vector<pair<long long, bool> > b, bb;
long long lft[528000], rght[528000];
int main() {
  int n;
  long long k, len = INF;
  scanf("%d %I64d", &n, &k);
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    scanf("%I64d %I64d", &a[i].first, &a[i].second);
    len = min(len, a[i].second - a[i].first);
  }
  lucky.reserve(529000);
  q.push(4), q.push(7);
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    lucky.push_back(x);
    if (x > INF2) continue;
    q.push(x * 10ll + 4ll), q.push(x * 10ll + 7ll);
  }
  for (int i = 0; i < lucky.size(); ++i) lft[i] = INF, rght[i] = INF;
  b.reserve(630000);
  bb.reserve(630000);
  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());
  inplace_merge(b.begin(), b.begin() + lucky.size(), b.end());
  long long cnt = 0, cost = 0;
  int lcnt = 0;
  if (!b[0].second)
    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)
      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());
  inplace_merge(b.begin(), b.begin() + lucky.size(), b.end());
  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)
      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] <= len &&
           lft[r] + rght[i] <= k)
      ans = max(ans, r - i + 1), r++;
  }
  printf("%d", ans);
  return 0;
}
