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 a, b, c, d, k;
vector<int> lucky;
void dfs(long long x) {
  if (x > 1000000000) return;
  lucky.push_back(x);
  dfs(x * 10 + 4), dfs(x * 10 + 7);
}
int inter(int a, int b, int c, int d) {
  a = max(a, c), b = min(b, d);
  return max(0, b - a + 1);
}
int main() {
  cin >> a >> b >> c >> d >> k;
  dfs(4), dfs(7);
  sort(lucky.begin(), lucky.end());
  long long good = 0;
  for (int i = 0; i < (int)lucky.size(); i++)
    if (inter(lucky[i], lucky[i], a, b))
      for (int j = 0; j < (int)lucky.size(); j++)
        if (inter(lucky[j], lucky[j], c, d)) good += abs(j - i) + 1 == k;
  lucky.insert(lucky.begin(), 0);
  lucky.push_back(1000000005);
  for (int i = 0; i < (int)lucky.size() - 1; i++)
    for (int j = 0; j < (int)lucky.size() - 1; j++)
      if (abs(i - j) == k)
        good += (long long)inter(lucky[i] + 1, lucky[i + 1] - 1, a, b) *
                (long long)inter(lucky[j] + 1, lucky[j + 1] - 1, c, d);
  for (int i = 1; i < (int)lucky.size() - 1; i++)
    if (inter(lucky[i], lucky[i], a, b))
      for (int j = 0; j < (int)lucky.size() - 1; j++) {
        long long add = inter(lucky[j] + 1, lucky[j + 1] - 1, c, d);
        if (j < i && i - j == k)
          good += add;
        else if (j >= i && j - i + 1 == k)
          good += add;
      }
  for (int i = 1; i < (int)lucky.size() - 1; i++)
    if (inter(lucky[i], lucky[i], c, d))
      for (int j = 0; j < (int)lucky.size() - 1; j++) {
        long long add = inter(lucky[j] + 1, lucky[j + 1] - 1, a, b);
        if (j < i && i - j == k)
          good += add;
        else if (j >= i && j - i + 1 == k)
          good += add;
      }
  printf("%.10lf\n",
         (double)good / ((long long)(b - a + 1) * (long long)(d - c + 1)));
  return 0;
}