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 int N = 2e5 + 7;
const int M = 257;
const int inf = 1e9 + 7;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-18;
vector<long long> b;
void go(long long x) {
  if (x >= base) return;
  if (x > 0) b.push_back(x);
  go(x * 10 + 4);
  go(x * 10 + 7);
}
long long l[N], r[N];
long long mindist = base + 1;
int n;
long long k;
int p = 0, q = 0;
long double sumL = 0, sumR = 0;
bool check(long long u, long long v) {
  if (v - u + 1 > mindist) return false;
  while (p < n && r[p] <= v) sumR += r[p++];
  while (q < n && l[q] <= u) sumL -= l[q++];
  long double moveR = (long double)v * p - sumR;
  long double moveL = sumL - (long double)u * (n - q);
  return moveL + moveR <= (long double)k;
}
void solve() {
  go(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", l + i, r + i);
    mindist = min(mindist, r[i] - l[i] + 1);
  }
  sort(l, l + n);
  sort(r, r + n);
  sort(b.begin(), b.end());
  for (int i = 0; i < n; i++) sumL += (long double)l[i];
  int j = 0;
  long long ans = 0;
  for (int i = 0; i < b.size(); i++) {
    if (j < i) j = i;
    while (j < b.size() && check(b[i], b[j])) j++;
    ans = max(ans, (long long)j - i);
  }
  cout << ans;
}
int main() { solve(); }
