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;
int digit[2] = {4, 7};
vector<int> num;
int p1, p2, v1, v2, k;
void ini(long long int x) {
  if (x * 10 > (10000 * 100000)) return;
  for (int i(0); i < 2; ++i) {
    num.push_back(x * 10 + digit[i]);
    ini(x * 10 + digit[i]);
  }
  return;
}
double get_num(int si, int di) {
  double ans = 0;
  if (num[si - 1] > p2 || num[di + 1] < v1 || num[si - 1] > v2) {
    ans = 0;
  } else {
    bool flag1 = false, flag2 = false;
    if (num[di] <= v2 && num[si] >= p1) {
      double sec2 = min(num[di + 1], v2), sec1 = max(num[di], v1);
      double fir1 = max(num[si - 1], p1), fir2 = min(num[si], p2);
      if (num[di + 1] > v2) sec2++;
      if (num[si - 1] < p1) fir1--;
      ans += (sec2 - sec1) * (fir2 - fir1);
      flag1 = true;
    }
    if (num[di] <= p2 && num[si] >= v1) {
      double sec2 = min(num[di + 1], p2), sec1 = num[di];
      double fir1 = max(num[si - 1], v1), fir2 = min(num[si], v2);
      if (num[si - 1] < v1) fir1--;
      if (num[di + 1] > p2) sec2++;
      ans += (sec2 - sec1) * (fir2 - fir1);
      flag2 = true;
    }
    if (num[di] == num[si] && (flag1 && flag2)) ans--;
  }
  return ans;
}
int main() {
  scanf("%d%d%d%d%d", &p1, &p2, &v1, &v2, &k);
  ini(0);
  num.push_back(-1);
  num.push_back(1000000009);
  sort(num.begin(), num.end());
  if (p1 > v1) {
    swap(p1, v1);
    swap(p2, v2);
  }
  int now(-1), end(-1);
  for (int i(0); i < num.size(); ++i) {
    if (num[i] > max(p2, v2)) {
      end = i;
      break;
    }
    if (now != -1) continue;
    if (num[i] <= max(p2, v2) && num[i] >= min(p1, v1)) {
      now = i;
    }
  }
  if (now == -1) {
    cout << 0.000000 << endl;
    return 0;
  } else {
    if (end - now < k) {
      cout << 0.0000000 << endl;
      return 0;
    }
  }
  double ans = 0;
  for (int i(now); i < end && i + k - 1 < end; ++i) {
    int j = i + k - 1;
    ans += get_num(i, j);
  }
  double div = (p2 - p1 + 1);
  div *= (v2 - v1 + 1);
  ans /= div;
  if (ans > 1) ans = 1;
  printf("%.9lf\n", ans);
  return 0;
}