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;
const bool local = false;
const int MAXK = 5000;
int a[MAXK];
inline long long intersect(int a, int b, int c, int d) {
  int ans = min(b, d) - max(a, c) + 1;
  return (ans > 0 ? ans : 0);
}
int main() {
  if (local) {
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  int n;
  int cnt = 0;
  a[cnt++] = 0;
  for (int i = 0; i < 10; i++) {
    if (!i) continue;
    for (int j = 0; j < (1 << i); j++) {
      for (int k = 0; k < i; k++) {
        if (j & (1 << (i - 1 - k)))
          a[cnt] = a[cnt] * 10 + 7;
        else
          a[cnt] = a[cnt] * 10 + 4;
      }
      cnt++;
    }
  }
  a[cnt++] = 1e9 + 1;
  int p1, p2, v1, v2, k;
  scanf("%d %d %d %d %d", &p1, &p2, &v1, &v2, &k);
  long long good = 0;
  for (int i = 0; i < cnt - k - 1; i++) {
    good += intersect(a[i] + 1, a[i + 1], p1, p2) *
            intersect(a[i + k], a[i + k + 1] - 1, v1, v2);
    good += intersect(a[i] + 1, a[i + 1], v1, v2) *
            intersect(a[i + k], a[i + k + 1] - 1, p1, p2);
  }
  if (k == 1) {
    for (int i = 0; i < cnt - 1; i++) {
      if (!i) continue;
      if ((a[i] >= max(p1, v1)) && (a[i] <= min(p2, v2))) good--;
    }
  }
  long double ans = good;
  ans /= (p2 - p1 + 1);
  ans /= (v2 - v1 + 1);
  printf("%.9f", (double)ans);
}