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 long long base = 10000000000000LL;
long long open[100005];
long long close[100005];
long long len = 2000000000000000000LL;
long long good[600000];
int q = 0;
void doit(long long sum, int left) {
  if (!left) {
    good[q++] = sum;
  } else {
    if (!sum) {
      doit(sum, left - 1);
    }
    doit(sum * 10 + 4, left - 1);
    doit(sum * 10 + 7, left - 1);
  }
}
int n;
long long k;
long long low = 0, high = 0;
void add(long long val) {
  high += val / base;
  low += val % base;
  if (low >= base) {
    ++high;
    low -= base;
  }
}
void add(long long val, int q) {
  long long to_high = val / base, to_low = val % base;
  high += to_high * q;
  low += to_low * q;
  high += low / base;
  low %= base;
}
void sub(long long val) {
  high -= val / base;
  low -= val % base;
  if (low < 0) {
    --high;
    low += base;
  }
}
void sub(long long val, int q) {
  long long to_high = val / base, to_low = val % base;
  high -= to_high * q;
  low -= to_low * q;
  if (low < 0) {
    long long q = (-low + base - 1) / base;
    low += q * base;
    high -= q;
  }
}
bool ok() {
  if (high < k / base || high == k / base && low <= k % base) {
    return true;
  } else {
    return false;
  }
}
bool check(int val) {
  int l = 0, r = 0;
  low = 0;
  high = 0;
  int to_left = n, to_right = 0;
  for (int i = 0; i < n; i++) {
    add(open[i]);
  }
  for (int i = 1; i + val - 1 < q; i++) {
    while (l < n && open[l] <= good[i]) {
      sub(open[l] - good[i - 1]);
      to_left--;
      ++l;
    }
    sub(good[i] - good[i - 1], to_left);
    add(good[i + val - 1] - good[i + val - 2], to_right);
    while (r < n && close[r] <= good[i + val - 1]) {
      add(good[i + val - 1] - close[r]);
      to_right++;
      ++r;
    }
    if (good[i + val - 1] - good[i] <= len && ok()) {
      return true;
    }
  }
  return false;
}
int main() {
  doit(0, 18);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> open[i] >> close[i];
    len = min(len, close[i] - open[i]);
  }
  sort(open, open + n);
  sort(close, close + n);
  int l = 0, r = q + 1;
  while (r - l > 1) {
    int mb = (l + r) >> 1;
    if (check(mb)) {
      l = mb;
    } else {
      r = mb;
    }
  }
  printf("%d\n", l);
  return 0;
}
