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;
vector<long long> num;
struct segment {
  long long l, r;
} a[600001];
long long costLeft[600001], costRight[600001];
int n, m, kq;
long long k, minLen;
void gen() {
  for (int len = 1; len <= 18; len++) {
    for (int mask = 0; mask < (1 << len); mask++) {
      long long cur = 0;
      for (int i = 0; i < len; i++)
        if ((mask >> i) & 1)
          cur = cur * 10 + 4;
        else
          cur = cur * 10 + 7;
      num.push_back(cur);
    }
  }
  sort(num.begin(), num.end());
  m = num.size();
}
bool cmp1(segment a, segment b) {
  if (a.r != b.r) return a.r < b.r;
  return a.l < b.l;
}
void processRight() {
  sort(a, a + n, cmp1);
  int cur = 0;
  for (int i = 0; i < m; i++) {
    if (i > 0)
      if (2e18 / (num[i] - num[i - 1]) > cur)
        costRight[i] = costRight[i - 1] + cur * (num[i] - num[i - 1]);
      else {
        costRight[i] = k + 1;
        continue;
      }
    if (costRight[i] > k) costRight[i] = k + 1;
    while (cur < n && a[cur].r <= num[i]) {
      costRight[i] += num[i] - a[cur].r;
      if (costRight[i] > k) costRight[i] = k + 1;
      cur++;
    }
  }
}
bool cmp2(segment a, segment b) {
  if (a.l != b.l) return a.l > b.l;
  return a.r > b.r;
}
void processLeft() {
  sort(a, a + n, cmp2);
  int cur = 0;
  for (int i = m - 1; i >= 0; i--) {
    if (i < m - 1)
      if (2e18 / (num[i + 1] - num[i]) > cur)
        costLeft[i] = costLeft[i + 1] + cur * (num[i + 1] - num[i]);
      else {
        costLeft[i] = k + 1;
        continue;
      }
    if (costLeft[i] > k) costLeft[i] = k + 1;
    while (cur < n && a[cur].l >= num[i]) {
      costLeft[i] += a[cur].l - num[i];
      if (costLeft[i] > k) costLeft[i] = k + 1;
      cur++;
    }
  }
}
int main() {
  scanf("%d%I64d", &n, &k);
  minLen = 1e18;
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &a[i].l, &a[i].r);
    minLen = min(minLen, a[i].r - a[i].l + 1);
  }
  gen();
  processLeft();
  processRight();
  int j = 0;
  for (int i = 0; i < m; i++) {
    while (j < m && costLeft[i] + costRight[j] <= k &&
           num[j] - num[i] + 1 <= minLen)
      j++;
    j = max(j, i);
    kq = max(kq, j - i);
  }
  cout << kq << endl;
  return 0;
}
