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;
const long long INF = (long long)(2e18);
long long n, k, Min;
long long l[100005], r[100005];
vector<long long> lucky;
long long L[600000], R[600000];
double sum[100005];
int main() {
  for (int i = 1; i <= 18; i++)
    for (int j = 0; j < (1 << i); j++) {
      long long cur = 0;
      for (int k = 0; k < i; k++)
        if (j & (1 << k))
          cur = cur * 10 + 7;
        else
          cur = cur * 10 + 4;
      lucky.push_back(cur);
    }
  sort(lucky.begin(), lucky.end());
  cin >> n >> k;
  int N = lucky.size();
  Min = INF;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    if (r[i] - l[i] < Min) Min = r[i] - l[i];
  }
  sort(l, l + n);
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i < n; i++)
    sum[i] = sum[i - 1] + (double)(l[i] - l[i - 1]) * (double)(n - i);
  for (int i = 0; i < N; i++) {
    int s = 0, t = n - 1;
    while (s < t) {
      int mid = (s + t) / 2;
      if (l[mid] <= lucky[i])
        s = mid + 1;
      else
        t = mid;
    }
    if (l[s] <= lucky[i]) s++;
    if (s > n - 1)
      L[i] = 0;
    else {
      double tmp =
          (double)(l[s] - lucky[i]) * (double)(n - s) + sum[n - 1] - sum[s];
      if (tmp >= INF)
        L[i] = INF;
      else
        L[i] = (long long)tmp;
    }
  }
  sort(r, r + n);
  memset(sum, 0, sizeof(sum));
  for (int i = n - 2; i >= 0; i--)
    sum[i] = sum[i + 1] + (double)(r[i + 1] - r[i]) * (double)(i + 1);
  for (int i = 0; i < N; i++) {
    int s = 0, t = n - 1;
    while (s < t) {
      int mid = (s + t) / 2;
      if (r[mid] >= lucky[i])
        t = mid;
      else
        s = mid + 1;
    }
    if (r[s] >= lucky[i]) s--;
    if (s < 0)
      R[i] = 0;
    else {
      double tmp =
          (double)(lucky[i] - r[s]) * (double)(s + 1) + sum[0] - sum[s];
      if (tmp >= INF)
        R[i] = INF;
      else
        R[i] = (long long)tmp;
    }
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    int s = 0, t = N - 1;
    while (s < t) {
      int mid = (s + t) / 2;
      if (lucky[mid] - lucky[i] > Min)
        t = mid;
      else
        s = mid + 1;
    }
    if (lucky[s] - lucky[i] > Min) s--;
    t = s;
    s = i;
    while (s < t) {
      int mid = (s + t) / 2;
      if (L[i] + R[mid] > k)
        t = mid;
      else
        s = mid + 1;
    }
    if (L[i] + R[s] > k) s--;
    if (L[i] + R[s] <= k && s - i + 1 > ans) ans = s - i + 1;
  }
  cout << ans << endl;
  return 0;
}
