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>
#pragma comment(linker, "/STACK:65777216")
using namespace std;
int n;
long long k;
vector<long long> v;
void go(long long c) {
  if (c) v.push_back(c);
  if (c <= 100000000000000000ll) {
    go(c * 10 + 4);
    go(c * 10 + 7);
  }
}
long long v1[1111111];
long long v2[1111111];
long long n1[1111111];
long long n2[1111111];
pair<long long, long long> m[111111];
long long diff;
bool good(int num) {
  for (int i(0); i < (v.size()); i++) {
    long long start = v[i];
    if (i + num - 1 >= v.size()) continue;
    long long end = v[i + num - 1];
    if (end - start > diff) continue;
    long long val = v1[i] + v2[i + num - 1];
    if (val <= k) return 1;
  }
  return 0;
}
int main() {
  go(0);
  sort((v).begin(), (v).end());
  cin >> n >> k;
  diff = 1000000000000000000ll;
  for (int i(0); i < (n); i++)
    scanf("%I64d %I64d", &m[i].first, &m[i].second),
        diff = min(diff, m[i].second - m[i].first);
  sort(m, m + n);
  int pos = n - 1;
  for (int i = v.size() - 1; i >= 0; i--) {
    if (v1[i + 1] + n1[i + 1] * double(v[i + 1] - v[i]) > 2 * k)
      v1[i] = k + 1;
    else
      v1[i] = v1[i + 1] + n1[i + 1] * (v[i + 1] - v[i]);
    v1[i] = min(v1[i], k + 1);
    n1[i] = n1[i + 1];
    while (pos >= 0 && m[pos].first >= v[i]) {
      n1[i]++;
      v1[i] += (m[pos].first - v[i]);
      v1[i] = min(v1[i], k + 1);
      pos--;
    }
  }
  for (int i(0); i < (n); i++) swap(m[i].first, m[i].second);
  sort(m, m + n);
  for (int i(0); i < (n); i++) swap(m[i].first, m[i].second);
  pos = 0;
  for (int i = 0; i < v.size(); i++) {
    if (i) {
      if (v2[i - 1] + n2[i - 1] * double(v[i] - v[i - 1]))
        v2[i] = k + 1;
      else
        v2[i] = v2[i - 1] + n2[i - 1] * (v[i] - v[i - 1]);
      v2[i] = min(v2[i], k + 1);
      n2[i] = n2[i - 1];
    }
    while (pos < n && m[pos].second <= v[i]) {
      n2[i]++;
      v2[i] += (v[i] - m[pos].second);
      v2[i] = min(v2[i], k + 1);
      pos++;
    }
  }
  int best = 0;
  int a = 1, b = v.size();
  while (a <= b) {
    int s = (a + b) >> 1;
    if (good(s))
      best = s, a = s + 1;
    else
      b = s - 1;
  }
  cout << best << endl;
  return 0;
}
