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> lucky;
void dfs(long long x) {
  if (x >= 1000000000) return;
  if (x) lucky.push_back(x);
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
long long Inter(long long x1, long long y1, long long x2, long long y2) {
  return max(min(y1, y2) - max(x1, x2) + 1, 0ll);
}
int main() {
  int i;
  dfs(0);
  lucky.push_back(0);
  lucky.push_back(1000000010);
  sort(lucky.begin(), lucky.end());
  long long pl, pr, vl, vr, k, X = 0, Y = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  Y = (vr - vl + 1) * (pr - pl + 1);
  for (i = 1; i + k < lucky.size(); i++) {
    X += Inter(pl, pr, lucky[i - 1] + 1, lucky[i]) *
         Inter(vl, vr, lucky[i + k - 1], lucky[i + k] - 1);
    X += Inter(vl, vr, lucky[i - 1] + 1, lucky[i]) *
         Inter(pl, pr, lucky[i + k - 1], lucky[i + k] - 1);
    if (k == 1)
      X -=
          Inter(vl, vr, lucky[i], lucky[i]) * Inter(pl, pr, lucky[i], lucky[i]);
  }
  printf("%.12lf\n", X * 1. / Y);
}