Problem: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.

Input

The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).

Output

On the single line print the result with an absolute error of no more than 10 - 9.

Examples

Input

1 10 1 10 2


Output

0.320000000000


Input

5 6 8 10 1


Output

1.000000000000

Note

Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>

In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.

In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int pl, pr, vl, vr, K;
int lucky_size;
int lucky[2333];
void get_lucky(int now) {
  lucky[++lucky_size] = now;
  if (now >= 100000000) return;
  get_lucky(now * 10 + 4);
  get_lucky(now * 10 + 7);
}
inline int __mymin(int a, int b) {
  if (a < b) return a;
  return b;
}
inline int __mymax(int a, int b) {
  if (a < b) return b;
  return a;
}
int main() {
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &K);
  get_lucky(0);
  sort(lucky + 1, lucky + lucky_size + 1);
  lucky[lucky_size + 1] = 1000000001;
  int i;
  int left, right;
  long long count = 0;
  long long temp = 0;
  for (i = 1; i + K <= lucky_size + 1; i++) {
    if (lucky[i] >= pl && lucky[i - 1] < pr && lucky[i + K] > vl &&
        lucky[i + K - 1] <= vr) {
      left = __mymax(pl, lucky[i - 1] + 1);
      right = __mymin(pr, lucky[i]);
      temp = (right - left + 1);
      left = __mymax(vl, lucky[i + K - 1]);
      if (right == left) count--;
      right = __mymin(vr, lucky[i + K] - 1);
      temp *= (long long)(right - left + 1);
      count += temp;
    }
    left = vl;
    vl = pl;
    pl = left;
    right = vr;
    vr = pr;
    pr = right;
    if (lucky[i] >= pl && lucky[i - 1] < pr && lucky[i + K] > vl &&
        lucky[i + K - 1] <= vr) {
      left = __mymax(pl, lucky[i - 1] + 1);
      right = __mymin(pr, lucky[i]);
      temp = (right - left + 1);
      left = __mymax(vl, lucky[i + K - 1]);
      right = __mymin(vr, lucky[i + K] - 1);
      temp *= (long long)(right - left + 1);
      count += temp;
    }
    left = vl;
    vl = pl;
    pl = left;
    right = vr;
    vr = pr;
    pr = right;
  }
  long long full = 0;
  full = (long long)(pr - pl + 1) * (long long)(vr - vl + 1);
  printf("%.15lf\n", count / (double)full);
  return 0;
}