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;
long long add(long long a, long long b) {
  assert(a >= 0 && b >= 0);
  return min(a + b, (1LL << 60));
}
long long mul(long long a, long long b) {
  assert(a >= 0 && b >= 0);
  long long c = a * b;
  if (a != 0 && (c % a || c / a != b)) return (1LL << 60);
  return min(c, (1LL << 60));
}
vector<long long> v, q0, q1;
long long a[100005], b[100005];
int n;
long long k, l;
vector<pair<long long, int> > e;
void gen(long long x) {
  if (x > 1000000000000000000LL) return;
  v.push_back(x);
  gen(10 * x + 4);
  gen(10 * x + 7);
}
int main() {
  gen(4);
  gen(7);
  sort(v.begin(), v.end());
  scanf("%d%lld", &n, &k);
  l = 1LL << 60;
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    scanf("%lld%lld", a + i, b + i), l = min(l, b[i] - a[i]);
  q0 = vector<long long>(v.size());
  q1 = vector<long long>(v.size());
  for (int i = 0, ThxDem = v.size(); i < ThxDem; ++i)
    e.push_back(make_pair(v[i], i));
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) e.push_back(make_pair(b[i], -1));
  sort(e.begin(), e.end());
  long long c = 0, lx = 0, q = 0;
  for (auto p : e) {
    q = add(q, mul(c, p.first - lx));
    lx = p.first;
    if (p.second < 0)
      c++;
    else
      q0[p.second] = q;
  }
  e.clear();
  for (int i = 0, ThxDem = v.size(); i < ThxDem; ++i)
    e.push_back(make_pair(v[i], i));
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) e.push_back(make_pair(a[i], -1));
  sort(e.begin(), e.end());
  reverse(e.begin(), e.end());
  c = 0;
  lx = (1LL << 60);
  q = 0;
  for (auto p : e) {
    q = add(q, mul(c, lx - p.first));
    lx = p.first;
    if (p.second < 0)
      c++;
    else
      q1[p.second] = q;
  }
  int j = 0, r = 0;
  for (int i = 0, ThxDem = v.size(); i < ThxDem; ++i) {
    if (j < i) j = i;
    while (j < v.size() && v[j] - v[i] <= l && add(q1[i], q0[j]) <= k) j++;
    r = max(r, j - i);
  }
  printf("%d\n", r);
  return 0;
}
