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>
T checkmax(T &a, T b) {
  return a = max(a, b);
}
template <class T>
T checkmin(T &a, T b) {
  return a = min(a, b);
}
const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = 0x3F3F3F3F;
const long long MX = 1000000000;
int lucky[1024], nl = 0;
void gen(long long seed) {
  if (seed >= MX) return;
  lucky[nl++] = seed;
  gen(seed * 10 + 4);
  gen(seed * 10 + 7);
}
int pl, pr, vl, vr;
long double lenp, lenv;
int k;
long long checkv(int id) {
  if (lucky[id] <= vl) return 0;
  int l = lucky[id - 1], r = lucky[id] - 1;
  checkmax(l, vl);
  checkmin(r, vr);
  if (l > r) return 0;
  return (r - l + 1);
}
long long check() {
  long long ans = 0;
  for (int i = 1; i < nl && lucky[i - 1] < pr; ++i) {
    if (lucky[i] < pl) continue;
    int l = lucky[i - 1] + 1, r = lucky[i];
    checkmax(l, pl);
    checkmin(r, pr);
    if (l > r) continue;
    long long p0 = (r - l + 1);
    long long pv = 0;
    if (i + k < nl && lucky[i + k - 1] <= vr) pv += checkv(i + k);
    ans += p0 * pv;
  }
  return ans;
}
double go() {
  long long ans = check();
  swap(pl, vl);
  swap(pr, vr);
  ans += check();
  if (k == 1) {
    for (int i = 1; i < nl && lucky[i] <= pr && lucky[i] <= vr; ++i) {
      if (lucky[i] >= pl && lucky[i] >= vl) --ans;
    }
  }
  return ans / lenp / lenv;
}
int main(int argc, char *argv[]) {
  gen(0);
  sort(lucky, lucky + nl);
  lucky[nl++] = MX + 1;
  while (cin >> pl >> pr >> vl >> vr >> k) {
    lenp = pr - pl + 1, lenv = vr - vl + 1;
    printf("%.10lf\n", go());
  }
}