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;
struct Tnode {
  int L, R;
};
const int maxn = 2007;
int K, n;
int Lucky[maxn];
Tnode A, B;
void dfs(int dep, int val) {
  if (dep > 10) return;
  if (val > 0) Lucky[++n] = val;
  dfs(dep + 1, val * 10 + 4);
  dfs(dep + 1, val * 10 + 7);
}
void prepare() {
  n = 0;
  Lucky[++n] = 0, Lucky[++n] = 1000000001;
  dfs(1, 0);
  sort(Lucky + 1, Lucky + 1 + n);
}
void work() {
  long long res = 0;
  for (int i = K + 2; i <= n; i++) {
    Tnode a, b, fir, lst;
    long long len1, len2;
    a.L = Lucky[i - K - 1] + 1, a.R = Lucky[i - K];
    b.L = Lucky[i - 1], b.R = Lucky[i] - 1;
    fir.L = max(a.L, A.L), fir.R = min(a.R, A.R);
    lst.L = max(b.L, B.L), lst.R = min(b.R, B.R);
    len1 = max(0, fir.R - fir.L + 1), len2 = max(0, lst.R - lst.L + 1);
    res += len1 * len2;
    fir.L = max(a.L, B.L), fir.R = min(a.R, B.R);
    lst.L = max(b.L, A.L), lst.R = min(b.R, A.R);
    len1 = max(0, fir.R - fir.L + 1), len2 = max(0, lst.R - lst.L + 1);
    res += len1 * len2;
    if (K == 1 && fir.R >= lst.L && fir.R >= fir.L && lst.R >= lst.L) res--;
  }
  long long len1 = A.R - A.L + 1, len2 = B.R - B.L + 1;
  long long sum = len1 * len2;
  double ans = (double)res / (double)sum;
  printf("%.10f\n", ans);
}
int main() {
  prepare();
  while (scanf("%d %d %d %d %d", &A.L, &A.R, &B.L, &B.R, &K) != EOF) work();
}