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<long long> lucky;
const long long inf = 1LL << 60;
int k;
long long pl, pr, vl, vr;
void dfs(long long x, int l) {
  lucky.push_back(x);
  if (l >= 11) return;
  dfs(10 * x + 4, l + 1);
  dfs(10 * x + 7, l + 1);
}
long long intersect(long long l1, long long r1, long long l2, long long r2) {
  long long ll = max(l1, l2), rr = min(r1, r2);
  if (ll > rr)
    return 0;
  else
    return (rr - ll + 1);
}
int main() {
  lucky.clear();
  dfs(4, 1);
  dfs(7, 1);
  sort(lucky.begin(), lucky.end());
  cin >> pl >> pr >> vl >> vr >> k;
  double a = 0, b = (double)(pr - pl + 1) * (vr - vl + 1);
  for (int i = 0; i < lucky.size(); i++) {
    int j = i + k - 1;
    if (j >= lucky.size()) break;
    long long l1 = (i == 0) ? -inf : lucky[i - 1] + 1;
    long long r1 = lucky[i];
    long long l2 = lucky[j];
    long long r2 = (j == lucky.size() - 1) ? inf : lucky[j + 1] - 1;
    a += (double)intersect(pl, pr, l1, r1) * intersect(vl, vr, l2, r2);
    a += (double)intersect(vl, vr, l1, r1) * intersect(pl, pr, l2, r2);
    if ((k == 1) && intersect(pl, pr, r1, r1) && intersect(vl, vr, l2, r2)) --a;
  }
  printf("%.18lf\n", a / b);
  return 0;
}