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> num;
long long n, pl, pr, vl, vr, k, ans;
void dfs(long long x) {
  if (x >= 1e9) return;
  num.push_back(x);
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
inline long long calc(long long l, long long r, long long x, long long y) {
  if (r < x || l > y) return 0;
  if (l > r || x > y) return 0;
  if (l >= x && l <= y) return min(r, y) - l + 1;
  if (r >= x && r <= y) return r - max(l, x) + 1;
  if (l < x && r > y) return y - x + 1;
}
int main() {
  dfs(0);
  n = num.size();
  num.push_back(1e16);
  sort(num.begin(), num.end());
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 1; i + k - 1 <= n; i++) {
    if (num[i] > vr && num[i] > pr) break;
    int j = i + k - 1;
    ans += calc(num[i - 1] + 1, num[i], pl, pr) *
           calc(num[j], num[j + 1] - 1, vl, vr);
    ans += calc(num[j], num[j + 1] - 1, pl, pr) *
           calc(num[i - 1] + 1, num[i], vl, vr);
    if (k == 1) {
      if (calc(pl, pr, num[i], num[i]) && calc(vl, vr, num[i], num[i])) ans--;
    }
  }
  printf("%.12lf", ans * 1.0 / (vr - vl + 1) / (pr - pl + 1), 1.0);
}