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;
long long p1, p2, v1, v2;
int K;
vector<long long> l;
double ret;
void rec(long long x) {
  if (x > 0) l.push_back(x);
  if ((long long)(x)*10 + 4 > 100000000005LL) return;
  rec(x * 10 + 4);
  rec(x * 10 + 7);
}
double countp(long long L, long long R) {
  if (L > R) return 0;
  long long A = max(p1, L), B = min(R, p2);
  if (A > B) return 0;
  return (B - A + 1) / (double)(p2 - p1 + 1);
}
double countv(long long L, long long R) {
  if (L > R) return 0;
  long long A = max(v1, L), B = min(R, v2);
  if (A > B) return 0;
  return (B - A + 1) / (double)(v2 - v1 + 1);
}
int main(void) {
  cin >> p1 >> p2 >> v1 >> v2 >> K;
  rec(0);
  sort(l.begin(), l.end());
  for (int i = 0; i + K - 1 < l.size(); ++i) {
    long long A = (i == 0) ? 1 : l[i - 1] + 1;
    long long B = l[i];
    long long C = l[i + K - 1];
    long long D = (i + K < l.size()) ? l[i + K] - 1 : max(p2, v2);
    ret += countp(A, B) * countv(C, D);
    ret += countp(C, D) * countv(A, B);
    if (B == C) ret -= countp(B, B) * countv(B, B);
  }
  printf("%.9lf\n", ret);
  return 0;
}