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 = 0x3f3f3f3f3f3f3f3fLL;
long long luck[1 << 19], nl;
void dfs(long long x) {
  if (x > 1000000000000000000LL) return;
  luck[nl++] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
const int Max = 100005;
long long le[Max], re[Max];
long long L[1 << 19], R[1 << 19];
bool overflow(long long a, long long b, long long c) {
  return (Inf - a) / c < b;
}
int main() {
  nl = 0;
  dfs(4);
  dfs(7);
  sort(luck, luck + nl);
  nl = unique(luck, luck + nl) - luck;
  int n;
  long long k;
  cin >> n >> k;
  long long gap = 1000000000000000000LL;
  for (int i = 0; i < n; ++i) {
    cin >> le[i] >> re[i];
    gap = min(gap, re[i] - le[i]);
  }
  sort(le, le + n);
  sort(re, re + n);
  {
    int i = nl - 1, j = n - 1;
    L[i] = 0;
    while (j >= 0 && le[j] > luck[i]) {
      L[i] += le[j] - luck[i];
      if (L[i] > Inf) L[i] = Inf;
      --j;
    }
    for (--i; i >= 0; --i) {
      if (L[i + 1] >= Inf ||
          overflow(L[i + 1], n - 1 - j, luck[i + 1] - luck[i]))
        L[i] = Inf;
      else
        L[i] = L[i + 1] + (n - 1 - j) * (luck[i + 1] - luck[i]);
      while (j >= 0 && le[j] > luck[i]) {
        L[i] += le[j] - luck[i];
        if (L[i] > Inf) L[i] = Inf;
        --j;
      }
    }
  }
  {
    int i = 0, j = 0;
    R[i] = 0;
    while (j < n && re[j] < luck[i]) {
      R[i] += luck[i] - re[j];
      if (R[i] > Inf) R[i] = Inf;
      ++j;
    }
    for (++i; i < nl; ++i) {
      if (R[i - 1] >= Inf || overflow(R[i - 1], j, luck[i] - luck[i - 1]))
        R[i] = Inf;
      else
        R[i] = R[i - 1] + j * (luck[i] - luck[i - 1]);
      while (j < n && re[j] < luck[i]) {
        R[i] += luck[i] - re[j];
        if (R[i] > Inf) R[i] = Inf;
        ++j;
      }
    }
  }
  int ans = 0;
  int lt = 0, rt = -1;
  while (lt < nl) {
    if (rt < lt - 1) rt = lt - 1;
    while (rt + 1 < nl && luck[rt + 1] - luck[lt] <= gap &&
           L[lt] + R[rt + 1] <= k)
      ++rt;
    ans = max(ans, rt + 1 - lt);
    ++lt;
  }
  printf("%d\n", ans);
  return 0;
}
