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;
#pragma comment(linker, "/STACK:160000000")
const int N = 9;
vector<int> a;
long long inter(int l, int r, int L, int R) {
  if (max(l, L) > min(r, R)) return 0;
  return min(r, R) - max(l, L) + 1;
}
int main() {
  int l, r, L, R, k;
  scanf("%d%d%d%d%d", &l, &r, &L, &R, &k);
  a.push_back(0);
  int l1 = 0;
  int r1 = 1;
  for (int i = 0; i < N; ++i) {
    for (int j = l1; j < r1; ++j) {
      a.push_back(a[j] * 10 + 4);
      a.push_back(a[j] * 10 + 7);
    }
    l1 = r1;
    r1 = a.size();
  }
  sort(a.begin(), a.end());
  a.push_back(1000000001);
  long long den = r - l + 1;
  den *= R - L + 1;
  long long num = 0;
  for (int i = 0; i + k + 1 < a.size(); ++i) {
    if (k == 1 && inter(a[i], a[i], l, r) == 1 &&
        inter(a[i], a[i], L, R) == 1) {
      --num;
    }
    num += inter(l, r, a[i] + 1, a[i + 1]) *
           inter(L, R, a[i + k], a[i + k + 1] - 1);
    num += inter(L, R, a[i] + 1, a[i + 1]) *
           inter(l, r, a[i + k], a[i + k + 1] - 1);
  }
  printf("%.10lf", (double)num / den);
  return 0;
}