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;
inline int read() {
  int res = 0;
  int neg;
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9' || ch == '-') {
      if (ch == '-')
        neg = -1;
      else
        neg = 1, res = ch - '0';
      break;
    }
  }
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9')
      res *= 10, res += ch - '0';
    else
      break;
  }
  return res * neg;
}
const int maxn = 100020;
const int maxm = 1000020;
const int MOd = 1e9 + 7;
int a;
long long K;
long double topf[maxn], tops[maxn];
map<long long, int> mp;
map<long long, int>::iterator it;
pair<long long, long long> ar[maxn];
int main() {
  scanf("%d %lld", &a, &K);
  long long mini = 1e18;
  for (int i = 1; i <= a; i++) {
    scanf("%lld %lld", &ar[i].first, &ar[i].second);
    mini = min(mini, (ar[i].second - ar[i].first));
  }
  sort(ar + 1, ar + 1 + a);
  vector<long long> w;
  for (int j = 1; j <= 18; j++) {
    for (int i = 0; i < (1 << j); i++) {
      long long c = 1;
      long long h = 0;
      for (int k = 0; k < j; k++, c *= 10) {
        if ((1 << k) & i)
          h += c * 7;
        else
          h += c * 4;
      }
      w.push_back(h);
    }
  }
  for (int i = 1; i <= a; i++) {
    topf[i] = topf[i - 1] + ar[i].first;
    tops[i] = tops[i - 1] + ar[i].second;
  }
  int ans = 0;
  for (int i = 0, j = 0, k = 1; i < w.size(); i++) {
    while (k <= a && ar[k].first <= w[i]) {
      mp[ar[k].second]++;
      k++;
    }
    long long t = K;
    if ((topf[a] - topf[k - 1]) - (long double)w[i] * (a - k + 1) > t) continue;
    t -= (topf[a] - topf[k - 1]) - (long double)w[i] * (a - k + 1);
    if (j < i) j = i;
    while (j < w.size() && w[j] <= w[i] + mini) {
    co:
      it = mp.begin();
      if (it != mp.end() && !it->second) {
        mp.erase(it);
        goto co;
      }
      if (it == mp.end() || it->first >= w[j]) {
        j++;
        continue;
      }
      if ((long double)(w[j] - it->first) * it->second > t) break;
      t -= (long double)(w[j] - it->first) * it->second;
      K -= (long double)(w[j] - it->first) * it->second;
      mp[w[j]] += it->second;
      mp.erase(it);
    }
    ans = max(ans, (j - i));
  }
  cout << ans << endl;
  return 0;
}
