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;
const int MX = (int)1e9 + 10000;
vector<long long> lucky;
void gen(long long c) {
  lucky.push_back(c);
  if (c * 10 + 4 < MX) gen(c * 10 + 4);
  if (c * 10 + 7 < MX) gen(c * 10 + 7);
}
double prob(long long a, long long b, long long c, long long d) {
  c = max(c, a);
  d = min(d, b);
  if (c > d) return 0;
  return (d - c + 1.0) / (b - a + 1.0);
}
void run() {
  lucky.clear();
  gen(0);
  lucky.push_back(MX);
  sort((lucky).begin(), (lucky).end());
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double ans = 0;
  int n = lucky.size();
  for (int i = 1; i < n - 1; i++) {
    ans +=
        prob(pl, pr, lucky[i], lucky[i]) *
        (k == 1 ? prob(vl, vr, lucky[i - 1] + 1, lucky[i + 1] - 1)
                : (i + k < n ? prob(vl, vr, lucky[i + k - 1], lucky[i + k] - 1)
                             : 0) +
                      (i >= k ? prob(vl, vr, lucky[i - k] + 1, lucky[i - k + 1])
                              : 0));
  }
  for (int i = 1; i < n; i++) {
    ans += prob(pl, pr, lucky[i - 1] + 1, lucky[i] - 1) *
           ((i + k < n ? prob(vl, vr, lucky[i + k - 1], lucky[i + k] - 1) : 0) +
            (i - k > 0 ? prob(vl, vr, lucky[i - k - 1] + 1, lucky[i - k]) : 0));
  }
  printf("%.9f\n", ans);
}
int main() { run(); }