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 MAXN = 110000;
const int MAXM = 610000;
const long long INF = 1000000000000000001LL;
long long k;
int n;
int m;
long long l[MAXN];
long long r[MAXN];
vector<long long> luck;
void gen(long long cur) {
  if (cur > INF) return;
  if (cur) luck.push_back(cur);
  gen(10 * cur + 4);
  gen(10 * cur + 7);
}
long long res[MAXM];
long long minlen;
void add(long long& a, long long b) {
  a += b;
  a = min(a, INF);
}
void mult(long long& a, long long b) {
  if (a < INF / b)
    a *= b;
  else
    a = INF;
}
bool check(int mid) {
  memset(res, 0, sizeof(res));
  long long cur = 0;
  int ptr = 0;
  for (int i = 0; i + mid <= m; i++) {
    long long temp = ptr;
    if (temp) mult(temp, luck[i + mid - 1] - luck[i + mid - 2]);
    add(cur, temp);
    while (ptr < n && r[ptr] < luck[i + mid - 1]) {
      add(cur, luck[i + mid - 1] - r[ptr]);
      ++ptr;
    }
    res[i] = cur;
  }
  cur = 0;
  ptr = 0;
  for (int i = m - mid; i >= 0; --i) {
    long long temp = ptr;
    if (temp) mult(temp, luck[i + 1] - luck[i]);
    add(cur, temp);
    while (ptr < n && l[ptr] > luck[i]) {
      add(cur, l[ptr] - luck[i]);
      ++ptr;
    }
    add(res[i], cur);
  }
  for (int i = 0; i + mid <= m; i++) {
    if (luck[i + mid - 1] - luck[i] <= minlen && res[i] <= k) {
      return true;
    }
  }
  return false;
}
int main() {
  minlen = INF;
  gen(0);
  sort(luck.begin(), luck.end());
  m = luck.size();
  scanf("%d", &n);
  scanf("%I64d", &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d %I64d", &l[i], &r[i]);
    minlen = min(minlen, r[i] - l[i]);
  }
  sort(l, l + n);
  reverse(l, l + n);
  sort(r, r + n);
  int lf = 0;
  int rg = luck.size() + 1;
  while (rg - lf > 1) {
    int mid = (lf + rg) / 2;
    if (check(mid))
      lf = mid;
    else
      rg = mid;
  }
  cout << lf << endl;
  return 0;
}
