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;
template <class T>
T sqr(T x) {
  return x * x;
}
double const pi = 3.1415926535897932384626433832795;
int const inf = (int)1e9 + 1;
long long const inf64 = (long long)4e18 + 1;
const string name = "b";
int l1, l2, r1, r2, k;
long long ans;
vector<int> now;
void rec(int zn, int n) {
  if (zn > 0) now.push_back(n);
  if (zn == 9) return;
  rec(zn + 1, n * 10 + 4);
  rec(zn + 1, n * 10 + 7);
}
int main() {
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  ans = 0;
  now.clear();
  rec(0, 0);
  now.push_back(0);
  now.push_back(inf);
  sort((now).begin(), (now).end());
  for (int i = 1; i < (int)now.size() - k; ++i) {
    int L1 = max(now[i - 1] + 1, l1);
    int R1 = min(now[i], r1);
    int L2 = max(now[i + k - 1], l2);
    int R2 = min(now[i + k] - 1, r2);
    if (L1 <= R1 && L2 <= R2)
      ans += (long long)(R1 - L1 + 1) * (long long)(R2 - L2 + 1);
  }
  for (int i = 1; i < (int)now.size() - k; ++i) {
    int L1 = max(now[i - 1] + 1, l2);
    int R1 = min(now[i], r2);
    int L2 = max(now[i + k - 1], l1);
    int R2 = min(now[i + k] - 1, r1);
    if (L1 <= R1 && L2 <= R2)
      ans += (long long)(R1 - L1 + 1) * (long long)(R2 - L2 + 1);
  }
  for (int i = 0; i < (int)(int)now.size(); i++)
    if (k == 1 && l1 <= now[i] && now[i] <= r1 && l2 <= now[i] && now[i] <= r2)
      ans--;
  printf("%.10lf\n",
         (double)ans / (double)(r2 - l2 + 1) / (double)(r1 - l1 + 1));
  return 0;
}