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;
static const double EPS = 1e-9;
static const double PI = acos(-1.0);
long long l[2];
long long r[2];
long long k;
long long seq[2000];
int m;
void dfs(long long num) {
  if (num > 1000000000) {
    return;
  }
  seq[m++] = num;
  dfs(num * 10 + 4);
  dfs(num * 10 + 7);
}
int main() {
  dfs(4);
  dfs(7);
  seq[m++] = 0;
  seq[m++] = 1000000000;
  sort(seq, seq + m);
  while (cin >> l[0] >> r[0] >> l[1] >> r[1] >> k) {
    long long all = (r[0] - l[0] + 1) * (r[1] - l[1] + 1);
    long long total = 0;
    int tail = 1;
    int head = k;
    while (head <= m) {
      for (int iter = 0; iter < (int)(2); iter++) {
        long long lv =
            min(seq[tail] + 1, r[iter] + 1) - max(seq[tail - 1] + 1, l[iter]);
        long long rv =
            min(seq[head + 1], r[iter ^ 1] + 1) - max(seq[head], l[iter ^ 1]);
        if (lv <= 0 || rv <= 0) {
          continue;
        }
        assert(lv >= 0);
        assert(rv >= 0);
        total += lv * rv;
      }
      if (k == 1 && l[0] <= seq[head] && seq[head] <= r[0] &&
          l[1] <= seq[head] && seq[head] <= r[1]) {
        total--;
      }
      head++;
      tail++;
    }
    printf("%.10f\n", total / (double)all);
  }
}