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;
pair<unsigned long long, unsigned long long> Seg[100009];
vector<unsigned long long> Luck;
unsigned long long N, K, MinSeg;
unsigned long long Left[1000000], Right[1000000];
void Prepare() {
  Luck.clear();
  for (int i = 1; i <= 18; i++)
    for (unsigned long long s = 0; s < (1 << i); s++) {
      unsigned long long num = 0;
      for (int l = 0; l < i; l++)
        if ((s >> l) & 1)
          num = num * 10 + 4;
        else
          num = num * 10 + 7;
      Luck.push_back(num);
    }
  sort(Luck.begin(), Luck.end());
}
void Init() {
  for (int i = 1; i <= N; i++) cin >> Seg[i].first >> Seg[i].second;
  MinSeg = Seg[1].second - Seg[1].first + 1;
  for (int i = 1; i <= N; i++)
    MinSeg = min(MinSeg, Seg[i].second - Seg[i].first + 1);
  unsigned long long Sum;
  for (int i = 0; i < Luck.size(); i++) Left[i] = K + 1, Right[i] = K + 1;
  sort(Seg + 1, Seg + N + 1);
  Sum = 0LL;
  for (int i = N, j = Luck.size() - 1; j >= 0; j--) {
    while (i > 0 && Seg[i].first >= Luck[j]) {
      Sum += Seg[i].first;
      i--;
    }
    Right[j] = Sum - (N - i) * Luck[j];
    if (Right[j] > K) break;
  }
  for (int i = 1; i <= N; i++) swap(Seg[i].first, Seg[i].second);
  sort(Seg + 1, Seg + N + 1);
  Sum = 0LL;
  for (int i = 1; i <= N; i++) swap(Seg[i].first, Seg[i].second);
  for (int i = 1, j = 0; j < Luck.size(); j++) {
    while (i <= N && Seg[i].second <= Luck[j]) {
      Sum += Seg[i].second;
      i++;
    }
    Left[j] = (i - 1) * Luck[j] - Sum;
    if (Left[j] > K) break;
  }
}
bool Check(unsigned long long Midans) {
  if (Midans == 0) return true;
  for (int i = 0; i + Midans - 1 < Luck.size(); i++) {
    if (Luck[i + Midans - 1] - Luck[i] + 1 > MinSeg) continue;
    if (Left[i + Midans - 1] + Right[i] <= K) return true;
  }
  return false;
}
void Solve() {
  unsigned long long Lans = 0, Rans = Luck.size();
  while (Lans < Rans) {
    unsigned long long Midans = (Lans + Rans + 1) / 2;
    bool pan = Check(Midans);
    if (pan)
      Lans = Midans;
    else
      Rans = Midans - 1;
  }
  cout << Lans << "\n";
}
int main() {
  cin >> N >> K;
  Prepare();
  Init();
  Solve();
}
