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>
int dx[] = {0, -1, 0, 1, -1, 1, -1, 1};
int dy[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int MOD = 1e8, N = 5e2 + 5, oo = 1e9;
const double pi = acos(-1);
using namespace std;
vector<long double> vec;
void solve(long double n) {
  if (n > 1e10) return;
  vec.push_back(n);
  solve(n * 10 + 7);
  solve(n * 10 + 4);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve(0);
  sort(vec.begin(), vec.end());
  vec.push_back(1e18);
  long double pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  long double ans = 0;
  for (int i = 1; i + k < (int)(vec).size(); i++) {
    long double a =
        max((long double)0, min(vec[i], pr) - max(vec[i - 1] + 1, pl) + 1);
    long double b = max((long double)0,
                        min(vec[i + k] - 1, vr) - max(vl, vec[i + k - 1]) + 1);
    ans += a * b;
    a = max((long double)0, min(vec[i], vr) - max(vec[i - 1] + 1, vl) + 1);
    b = max((long double)0,
            min(vec[i + k] - 1, pr) - max(pl, vec[i + k - 1]) + 1);
    ans += a * b;
    ans -= (k == 1 and pl <= vec[i] and vl <= vec[i] and vec[i] <= vr and
            vec[i] <= pr);
  }
  long double total = (pr - pl + 1) * (vr - vl + 1);
  cout << fixed << setprecision(12) << ans / total;
  return 0;
}