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;
vector<int> A;
int intersect(int a, int b, int c, int d, int niv) {
  if (niv == 2) return 0;
  if (c >= a && c <= b && d >= a && d <= b) return d - c + 1;
  if (c >= a && c <= b) return b - c + 1;
  if (d >= a && d <= b) return d - a + 1;
  return intersect(c, d, a, b, niv + 1);
}
void gen(int x) {
  if ((long long)x * 10 + 4 <= 1000000000)
    A.push_back(x * 10 + 4), gen(x * 10 + 4);
  if ((long long)x * 10 + 7 <= 1000000000)
    A.push_back(x * 10 + 7), gen(x * 10 + 7);
}
int pL, pR, vL, vR, K;
int main(void) {
  cin >> pL >> pR >> vL >> vR >> K;
  gen(0);
  sort(A.begin(), A.end());
  int i, j, k;
  double res = 0.0;
  for (i = 0; i + K - 1 < A.size(); i++) {
    int p, v;
    p = intersect(pL, pR, i == 0 ? 1 : A[i - 1] + 1, A[i], 0);
    v = intersect(vL, vR, A[i + K - 1],
                  i + K == A.size() ? 1000000000 : A[i + K] - 1, 0);
    res += (double)p / (pR - pL + 1) * (double)v / (vR - vL + 1);
    p = intersect(vL, vR, i == 0 ? 1 : A[i - 1] + 1, A[i], 0);
    v = intersect(pL, pR, A[i + K - 1],
                  i + K == A.size() ? 1000000000 : A[i + K] - 1, 0);
    res += (double)p / (vR - vL + 1) * (double)v / (pR - pL + 1);
    if (K == 1 && pL <= A[i] && pR >= A[i] && vL <= A[i] && vR >= A[i])
      res -= (double)1.0 / (pR - pL + 1) * (double)1.0 / (vR - vL + 1);
  }
  printf("%.12lf\n", res);
  return 0;
}