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 MAXL = 19;
const int MAXN = 100010;
const int MAXM = (1 << MAXL);
int n;
long long k;
long long inv_len;
vector<long long> lucknum;
long long lsum[MAXM], rsum[MAXM];
long long L[MAXN], R[MAXN];
void init() {
  scanf("%d%lld", &n, &k);
  inv_len = (1ll << 62);
  long long l, r;
  for (int i = 0; i < n; ++i) {
    scanf("%lld%lld", &l, &r);
    inv_len = min(inv_len, r - l);
    R[i] = r;
    L[i] = l;
  }
}
void lucknum_cal() {
  lucknum.clear();
  for (int len = 1; len < MAXL; ++len)
    for (int limit = (1 << len), num = 0; num < limit; ++num) {
      long long cur = 0;
      for (int l = len - 1; l >= 0; --l)
        cur = cur * 10 + ((num & (1 << l)) ? 7 : 4);
      lucknum.push_back(cur);
    }
}
void solve() {
  lucknum_cal();
  sort(L, L + n);
  sort(R, R + n);
  long long cur_sum = 0;
  int pt = 0;
  for (int i = 0; i < lucknum.size(); ++i) {
    for (; pt < n && R[pt] < lucknum[i]; ++pt) cur_sum += R[pt];
    if (i > 0 && rsum[i - 1] > k)
      rsum[i] = k + 1;
    else {
      rsum[i] = lucknum[i] * pt - cur_sum;
      if (rsum[i] > k) rsum[i] = k + 1;
    }
  }
  cur_sum = 0;
  pt = n - 1;
  for (int i = lucknum.size() - 1; i >= 0; --i) {
    for (; pt >= 0 && L[pt] > lucknum[i]; --pt) cur_sum += L[pt];
    if (i < lucknum.size() && lsum[i + 1] > k)
      lsum[i] = k + 1;
    else {
      lsum[i] = cur_sum - lucknum[i] * (n - 1 - pt);
      if (lsum[i] > k) lsum[i] = k + 1;
    }
  }
  int ans = 0, lf = 0;
  long long remain;
  for (int head = 0, tail = 0; tail < lucknum.size(); ++tail) {
    for (; lucknum[tail] - lucknum[head] > inv_len; ++head)
      ;
    remain = k - rsum[tail];
    for (; lf < lucknum.size() && lsum[lf] > remain; ++lf)
      ;
    ans = max(ans, tail - max(head, lf) + 1);
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
