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> lst;
void gen(long long n) {
  if (n > 1000000000) return;
  if (n > 0) lst.push_back(n);
  gen(n * 10 + 4);
  gen(n * 10 + 7);
}
int L1, R1, L2, R2, k;
double f(double x, double y, double L, double R) {
  return max(0.0, min(R, y) - max(L, x) + 1);
}
double cal(int x1, int y1, int x2, int y2) {
  double ret = 0;
  ret += f(x1, y1, L1, R1) * f(x2, y2, L2, R2);
  ret += f(x2, y2, L1, R1) * f(x1, y1, L2, R2);
  if (y1 == x2) ret -= f(y1, y1, L1, R1) * f(x2, x2, L2, R2);
  return ret;
}
int main() {
  gen(0);
  lst.push_back(-1);
  lst.push_back(1000000001);
  sort(lst.begin(), lst.end());
  double ans = 0;
  scanf("%d%d%d%d%d", &L1, &R1, &L2, &R2, &k);
  for (int i = 1; i + k < (int)lst.size(); i++)
    ans += cal(lst[i - 1] + 1, lst[i], lst[i + k - 1], lst[i + k] - 1);
  printf("%.14f\n", ans / (R1 - L1 + 1) / (R2 - L2 + 1));
  return 0;
}