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 rev(long long t) {
  long long res = 0;
  while (t) {
    if (t % 2)
      res *= 10, res += 7;
    else
      res *= 10, res += 4;
    t /= 10;
  }
  return res;
}
long long conv(long long t, int len) {
  long long res = 0, ans = 0, k = 0;
  while (t) {
    k++;
    if (t % 2)
      res *= 10, res += 7;
    else
      res *= 10, res += 4;
    t /= 2;
  }
  res = rev(res);
  for (int i = 0; i < len - k; i++) ans *= 10, ans += 4;
  for (int i = 0; i < k; i++) ans *= 10;
  ans += res;
  return ans;
}
const int MAXN = 1e5 + 10;
long double ens[MAXN], sts[MAXN], pse[MAXN], pss[MAXN];
vector<long double> num;
long double mn = 1e18, k;
int ans, n;
bool check(long double l, long double r) {
  long double res = 0;
  int t = upper_bound(ens, ens + n, r) - ens;
  t--;
  if (t >= 0) res -= pse[t], res += r * (t + 1);
  t = lower_bound(sts, sts + n, l) - sts;
  res += pss[t];
  res -= l * (n - t);
  return res <= k;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long double l, r;
    cin >> l >> r;
    mn = min(r - l, mn);
    ens[i] = r;
    sts[i] = l;
  }
  sort(ens, ens + n);
  pse[0] = ens[0];
  for (int i = 1; i < n; i++) pse[i] += pse[i - 1] + ens[i];
  sort(sts, sts + n);
  pss[n - 1] = sts[n - 1];
  for (int i = n - 2; i >= 0; i--) pss[i] += pss[i + 1] + sts[i];
  bool R = 0;
  for (int i = 1; i < 30; i++) {
    if (R) break;
    for (int j = 0; j < (1 << i); j++) {
      long long t = conv(j, i);
      if (t > 1e18) {
        R = 1;
        break;
      }
      num.push_back(t);
    }
  }
  for (int i = 0; i < num.size(); i++) {
    long long s = 0, e = mn + 1, r = num[i], l;
    if (!check(r, r)) continue;
    while (s + 1 != e) {
      long long mid = (s + e) / 2;
      l = r - mid;
      if (check(l, r))
        s = mid;
      else
        e = mid;
    }
    ans = max(ans,
              i -
                  (int)(lower_bound(num.begin(), num.begin() + i + 1, r - s) -
                        num.begin()) +
                  1);
  }
  cout << ans;
}
