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 MAXN = 111111;
const int oo = 0X1FFFFFFF;
int PP = 1000000007;
const long double PI = 3.141592653589793;
template <typename TT>
void read(TT &x) {
  char ch;
  for (ch = getchar(); ch > '9' || ch < '0'; ch = getchar())
    ;
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
}
int n, m, a[2222], b[2222], aa[2222], bb[2222], pnt = 0;
set<pair<long long, int> > p;
set<pair<long long, int> >::iterator it, is, ii, jj;
void work(int l, int r, int *a) {
  memset(a, 0, sizeof(a));
  it = p.lower_bound(make_pair(l, oo));
  if (it->first > l) it--;
  is = p.upper_bound(make_pair(r, 0));
  if (is->first <= r) is++;
  ii = it;
  ii++;
  if (ii == is) {
    a[it->second] = r - l + 1;
    return;
  }
  ii = it;
  for (jj = ii, jj++; jj != is; ii = jj, jj++) {
    a[ii->second] = jj->first - ii->first;
  }
  ii = it, ii++;
  a[it->second] = ii->first - l;
  ii = is, ii--;
  a[ii->second] = r - ii->first + 1;
}
int calc(int l, int r) {
  if (l > r) return 0;
  int x = 0;
  for (it = p.begin(); it != p.end(); it++) {
    if (it->first >= l && it->first <= r) x++;
  }
  return x;
}
int k;
void Init() {
  int l, r, x;
  p.insert(make_pair(0, 0));
  for (int i = 1; i <= 9; i++)
    for (int j = 0; j < 1 << i; j++) {
      x = j;
      int u = 0;
      for (int k = i - 1; k >= 0; k--) u *= 10, u += ((1 << k) & (x)) ? 7 : 4;
      p.insert(make_pair(u, ++pnt));
    }
  long long ans1;
  int ll, rr;
  {
    read(l);
    read(r);
  };
  work(l, r, a);
  work(l - 1, r - 1, aa);
  ans1 = (r - l + 1);
  ll = l, rr = r;
  {
    read(l);
    read(r);
  };
  work(l, r, b);
  work(l - 1, r - 1, bb);
  ans1 *= 1LL * (r - l + 1);
  read(k);
  long long ans = 0;
  for (int i = 0; i <= pnt - k; i++)
    ans += 1LL * aa[i] * b[i + k] + 1LL * bb[i] * a[i + k];
  if (k == 1) {
    ans -= calc(max(ll, l), min(rr, r));
  }
  printf("%.9lf", (double)ans / ans1);
}
int main() { Init(); }