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 = 1 << 29;
const double EPS = 1e-10;
const double PI = 3.1415926535897932384626433832795;
const int CLK = CLOCKS_PER_SEC;
vector<long long> lnums;
const long long lm = 1000000000;
void dfs(long long v) {
  long long v1 = v * 10 + 4;
  long long v2 = v * 10 + 7;
  if (v1 <= lm) {
    lnums.push_back(v1);
    dfs(v1);
  }
  if (v2 <= lm) {
    lnums.push_back(v2);
    dfs(v2);
  }
}
int main() {
  long long pl, pr, vl, vr;
  int k;
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0);
  lnums.push_back(0);
  lnums.push_back(lm + 1LL);
  sort(lnums.begin(), lnums.end());
  int sz = lnums.size();
  double res = 0.0;
  for (int i = 1; i + k < sz; i++) {
    long long ar = lnums[i];
    long long al = lnums[i - 1];
    long long br = lnums[i + k];
    long long bl = lnums[i + k - 1];
    double p1 = (double)max(min(ar, pr) - max(al + 1, pl) + 1LL, 0LL) *
                max(min(br - 1, vr) - max(bl, vl) + 1LL, 0LL);
    double p2 = (double)max(min(br - 1, pr) - max(bl, pl) + 1LL, 0LL) *
                max(min(ar, vr) - max(al + 1, vl) + 1LL, 0LL);
    res += p1 + p2;
    if (k == 1 && pl <= ar && ar <= pr && vl <= ar && ar <= vr) res -= 1;
  }
  cout << fixed << setprecision(12) << res / ((vr - vl + 1) * (pr - pl + 1))
       << endl;
  return 0;
}