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>
#pragma comment(linker, "/STACK:60000000")
using namespace std;
const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
const string task = "";
template <class T>
T sqr(T x) {
  return x * x;
}
long long pl, pr;
long long vl, vr;
int k;
vector<long long> A;
void go(long long v) {
  if (v > 1000000000LL) return;
  A.push_back(v);
  go(v * 10 + 4);
  go(v * 10 + 7);
}
long long calc(long long l1, long long r1, long long vl, long long vr) {
  long long l = max(l1, vl);
  long long r = min(r1, vr);
  if (l <= r) return r - l + 1;
  return 0;
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  A.clear();
  go(4);
  go(7);
  A.push_back(0);
  A.push_back(1000000001);
  sort(A.begin(), A.end());
  int l = 1;
  int r = l + k;
  long long res = 0;
  long long sum = (pr - pl + 1) * (vr - vl + 1);
  while (r < (int)A.size()) {
    int l1 = A[l - 1] + 1;
    int r1 = A[l];
    int l2 = A[r - 1];
    int r2 = A[r] - 1;
    res += calc(l1, r1, pl, pr) * calc(l2, r2, vl, vr);
    res += calc(l2, r2, pl, pr) * calc(l1, r1, vl, vr);
    if (r1 == l2 && (pl <= r1 && r1 <= pr && vl <= r1 && r1 <= vr)) res--;
    l++;
    r++;
  }
  printf("%.10lf\n", (double)res / (double)sum);
  return 0;
}