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;
int n;
long long k, mxd;
vector<long long> a, b;
vector<long long> luk;
void solve() {
  long long disL = 0, disR = 0, nL = n, nR = 0;
  int ans = 0;
  double dL = 0, dR = 0;
  for (int i = 0; i < n; i++) dL += a[i], disL += a[i];
  for (int i = 1, j = 1, pa = 0, pb = 0; i < (int)luk.size(); i++) {
    while (pa < n && a[pa] <= luk[i]) {
      --nL;
      disL -= a[pa];
      dL -= a[pa];
      pa++;
    }
    while (j < (int)luk.size() && luk[j] - luk[i] <= mxd) {
      if (dL - nL * 1.0 * luk[i] + nR * 1.0 * luk[j] - dR <= 1e19) {
        if (disL - nL * luk[i] + nR * luk[j] - disR <= k) {
          if (i <= j) {
            ans = max(ans, j - i + 1);
          }
        } else
          break;
      } else
        break;
      j++;
      while (pb < n && b[pb] <= luk[j]) {
        ++nR;
        disR += b[pb];
        dR += b[pb];
        pb++;
      }
    }
  }
  printf("%d\n", ans);
}
int main(void) {
  scanf("%d%I64d", &n, &k);
  mxd = (1LL << 60);
  for (int i = 0; i < n; i++) {
    long long x, y;
    scanf("%I64d%I64d", &x, &y);
    mxd = min(mxd, y - x);
    a.push_back(x);
    b.push_back(y);
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  luk.push_back(0);
  for (int i = 0; luk[i] * 10.0 < 1e18; i++) {
    luk.push_back(luk[i] * 10 + 4);
    luk.push_back(luk[i] * 10 + 7);
  }
  solve();
  return 0;
}
