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;
template <class T>
using vec = vector<T>;
const long long inf = numeric_limits<long long>::max() >> 1;
const long long infll = numeric_limits<long long>::max() >> 1;
const long long N = 1e5 + 1;
long long pl, pr, vl, vr, k;
long long ans;
vec<long long> lucky;
void input() { cin >> pl >> pr >> vl >> vr >> k; }
void generate(long long d, long long i) {
  if (i == 10) {
    return;
  }
  lucky.push_back(d);
  generate(10 * d + 4, i + 1);
  generate(10 * d + 7, i + 1);
}
void solve() {
  generate(0, 0);
  sort((lucky).begin(), (lucky).end());
  lucky.push_back(1e9);
  for (long long i = 1; i + k < (long long)((lucky).size()); ++i) {
    long long lx = max(pl, lucky[i - 1] + 1);
    long long ly = min(pr, lucky[i]);
    long long rx = max(vl, lucky[i + k - 1]);
    long long ry = min(vr, lucky[i + k] - 1);
    long long l = max(0ll, ly - lx + 1);
    long long r = max(0ll, ry - rx + 1);
    ans += l * r;
  }
  for (long long i = 1; i + k < (long long)((lucky).size()); ++i) {
    long long lx = max(vl, lucky[i - 1] + 1);
    long long ly = min(vr, lucky[i]);
    long long rx = max(pl, lucky[i + k - 1]);
    long long ry = min(pr, lucky[i + k] - 1);
    long long l = max(0ll, ly - lx + 1);
    long long r = max(0ll, ry - rx + 1);
    ans += l * r;
  }
  if (k == 1) {
    for (long long i = 1; i < (long long)((lucky).size()) - 1; ++i) {
      if (lucky[i] >= pl && lucky[i] <= pr && lucky[i] >= vl &&
          lucky[i] <= vr) {
        ans--;
      }
    }
  }
  cout << fixed << setprecision(9)
       << (double)ans / ((pr - pl + 1) * (vr - vl + 1)) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  input();
  solve();
}