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 int INF = 1000000007;
const int N = 100100;
int lucky[] = {4, 7};
int n, k;
long long p1, p2, v1, v2, a[N], f1, f2, l1, l2, ans;
void attemt(int id) {
  if (id == 10) return;
  for (int i = 0; i < 2; i++) {
    ans = ans * 10 + lucky[i];
    a[++n] = ans;
    attemt(id + 1);
    ans /= 10;
  }
}
long long process(int id) {
  int v;
  long long ans = 0;
  a[0] = -1;
  a[n + 1] = INF;
  for (int i = 1; i <= n; i++) {
    v = i + k - 1;
    if (v > n) continue;
    f1 = max(a[i - 1] + 1, p1);
    l1 = min(a[i], p2);
    f2 = max(a[v], v1);
    l2 = min(a[v + 1] - 1, v2);
    if (f1 <= l1 && f2 <= l2) {
      ans += (l1 - f1 + 1) * (l2 - f2 + 1);
      if (id == 2 && v == i) ans--;
    }
  }
  return ans;
}
int main() {
  attemt(1);
  sort(a + 1, a + n + 1);
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  long long sum = (long long)(p2 - p1 + 1) * (v2 - v1 + 1);
  long long l = process(1);
  swap(p1, v1);
  swap(p2, v2);
  l += process(2);
  printf("%.17lf", (double)l / sum);
}