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 long long MAXI = 2000000000ll;
vector<long long> vall;
set<long long> sall;
long long curr = 0;
void gen() {
  if (curr > MAXI) {
    return;
  }
  if (curr != 0) {
    sall.insert(curr);
  }
  curr *= 10;
  curr += 4;
  gen();
  curr -= 4;
  curr += 7;
  gen();
  curr -= 7;
  curr /= 10;
}
long long a1, a2, b1, b2, K;
int main() {
  gen();
  vall.assign(sall.begin(), sall.end());
  cin >> a1 >> a2 >> b1 >> b2 >> K;
  long long sumz = 0;
  for (int off = 0; off + K <= vall.size(); ++off) {
    long long prev = 0;
    long long first = vall[off];
    long long last = vall[off + K - 1];
    long long next = vall[off + K];
    if (off > 0) {
      prev = vall[off - 1];
    }
    if (first > a2 && first > b2) break;
    if (a1 <= first && b2 >= last) {
      long long xa1 = max(prev + 1, a1);
      long long xa2 = min(first, a2);
      long long xb1 = max(last, b1);
      long long xb2 = min(next - 1, b2);
      if (xa1 <= xa2 && xb1 <= xb2) {
        long long z = (xa2 - xa1 + 1) * (xb2 - xb1 + 1);
        sumz += z;
      }
    }
    if (b1 <= first && a2 >= last) {
      long long xb1 = max(prev + 1, b1);
      long long xb2 = min(first, b2);
      long long xa1 = max(last, a1);
      long long xa2 = min(next - 1, a2);
      if (xa1 <= xa2 && xb1 <= xb2) {
        long long z = (xb2 - xb1 + 1) * (xa2 - xa1 + 1);
        if (K == 1) {
          if (a1 <= first && last <= a2 && b1 <= first && last <= b2) {
            --z;
          }
        }
        sumz += z;
      }
    }
  }
  long long denom = (a2 - a1 + 1) * (b2 - b1 + 1);
  long double ans = ((long double)sumz) / ((long double)denom);
  cout.precision(12);
  cout.setf(ios::fixed);
  cout << ans << endl;
  return 0;
}