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 INF = 99999999;
long long pl, pr, vl, vr, k;
vector<long long> v;
void dfs(long long x, int digit) {
  if (digit >= 11) return;
  v.push_back(x);
  dfs(x * 10 + 4, digit + 1);
  dfs(x * 10 + 7, digit + 1);
}
inline long long intersection(long long l1, long long r1, long long l2,
                              long long r2) {
  long long l = max(l1, l2), r = min(r1, r2);
  return l > r ? 0 : r - l + 1;
}
int main() {
  v.clear();
  dfs(4, 1);
  dfs(7, 1);
  sort(v.begin(), v.end());
  while (cin >> pl >> pr >> vl >> vr >> k) {
    double cnt = 0, sum = (double)(pr - pl + 1) * (vr - vl + 1);
    for (int i = 0; i < v.size(); i++) {
      int j = i + k - 1;
      if (j >= v.size()) break;
      long long l1 = i == 0 ? -INF : v[i - 1] + 1;
      long long r1 = v[i];
      long long l2 = v[j];
      long long r2 = j >= v.size() - 1 ? INF : v[j + 1] - 1;
      cnt += intersection(l1, r1, pl, pr) * intersection(l2, r2, vl, vr);
      cnt += intersection(l1, r1, vl, vr) * intersection(l2, r2, pl, pr);
      if (k == 1 && intersection(r1, r1, pl, pr) &&
          intersection(l2, l2, vl, vr))
        cnt--;
    }
    printf("%.12lf\n", cnt / sum);
  }
  return 0;
}