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 long long oo = 1e9;
vector<long long> lucky;
void dfs(long long x) {
  if (x > oo) return;
  lucky.push_back(x);
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
long long overlap(long long a, long long b, long long x, long long y) {
  long long l = max(a, x);
  long long r = min(b, y);
  if (r < l) return 0;
  return r - l + 1;
}
long long pl, pr, vl, vr, k;
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0);
  sort(lucky.begin(), lucky.end());
  int n = lucky.size();
  lucky.push_back(oo);
  double ans = 0;
  for (int i = 1; i + k - 1 < n; i++) {
    long long ll = lucky[i - 1] + 1, lr = lucky[i];
    long long rl = lucky[i + k - 1], rr = lucky[i + k] - 1;
    long long p1 = overlap(pl, pr, ll, lr) * (long long)overlap(vl, vr, rl, rr);
    long long p2 = overlap(pl, pr, rl, rr) * (long long)overlap(vl, vr, ll, lr);
    ans += p1 + p2;
    if (p1 && p2) ans -= (k == 1);
  }
  printf("%.12lf\n", ans / double(pr - pl + 1) / double(vr - vl + 1));
  return 0;
}