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<int> vct;
void DFS(int d, int num) {
  if (d == 9) {
    return;
  } else {
    int a = num * 10 + 4;
    vct.push_back(a);
    DFS(d + 1, a);
    int b = num * 10 + 7;
    vct.push_back(b);
    DFS(d + 1, b);
  }
}
long long calc(int a, int b, int l, int r) {
  int x = max(a, l);
  int y = min(b, r);
  if (x > y) {
    return 0;
  } else {
    return y - x + 1;
  }
}
int main() {
  vct.clear();
  DFS(0, 0);
  vct.push_back(-1);
  vct.push_back(1000000009);
  sort(vct.begin(), vct.end());
  int ss = vct.size();
  int p1, p2, v1, v2;
  int k;
  while (scanf("%d %d %d %d %d", &p1, &p2, &v1, &v2, &k) != EOF) {
    long long ans = 0;
    for (int i = 1; i + k < ss; i++) {
      int x = vct[i - 1] + 1;
      int a = vct[i];
      int b = vct[i + k - 1];
      int y = vct[i + k] - 1;
      long long tmp = 0;
      tmp += calc(x, a, p1, p2) * calc(b, y, v1, v2);
      tmp += calc(x, a, v1, v2) * calc(b, y, p1, p2);
      if (k == 1 && v1 <= a && a <= p2 && tmp > 0) {
        tmp--;
      }
      ans += tmp;
    }
    double a = 1.0 * ans / (p2 - p1 + 1) / (v2 - v1 + 1);
    printf("%.20f\n", a);
  }
  return 0;
}