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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
const long long MAX = 1e17;
int n, u = 0;
long long m;
long long l[100001], r[100001];
long long cl[1000001], cr[1000001];
long long sl[1000001], sr[1000001];
long long md;
long long v[1000000];
int go(long long x) {
  if (x > 0) v[u++] = x;
  if (x <= MAX) {
    go(x * 10 + 4);
    go(x * 10 + 7);
  }
  return 0;
}
long long calc(long long a, long long b) {
  if (a == 0 || b == 0) return 0;
  if (a > m / b) return m + 1;
  return a * b;
}
int main() {
  cin >> n >> m;
  md = 1e18;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    md = min(md, r[i] - l[i]);
  }
  go(0);
  sort(v, v + u);
  sort(l, l + n);
  sort(r, r + n);
  reverse(r, r + n);
  sl[n - 1] = sr[n - 1] = 0;
  for (int i = n - 2; i >= 0; i--) {
    sl[i] = min(sl[i + 1] + calc(n - i - 1, l[i + 1] - l[i]), m + 1);
    sr[i] = min(sr[i + 1] + calc(n - i - 1, r[i] - r[i + 1]), m + 1);
  }
  int x = n;
  for (int i = u - 1; i >= 0; i--) {
    while (x > 0 && l[x - 1] >= v[i]) x--;
    if (x < n) cl[i] = min(sl[x] + calc(n - x, l[x] - v[i]), m + 1);
  }
  x = n;
  for (int i = 0; i < u; i++) {
    while (x > 0 && r[x - 1] <= v[i]) x--;
    if (x < n) cr[i] = min(sr[x] + calc(n - x, v[i] - r[x]), m + 1);
  }
  x = u - 1;
  int y = u - 1;
  int ans = 0;
  while (x >= 0) {
    while (y >= x && (v[y] - v[x] > md || cl[x] + cr[y] > m)) y--;
    if (x <= y) ans = max(ans, y - x + 1);
    x--;
  }
  printf("%d\n", ans);
  return 0;
}
