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> v;
void dfs(int i, long long num) {
  if (i == 10) return;
  dfs(i + 1, num * 10 + 4);
  dfs(i + 1, num * 10 + 7);
  v.push_back(num * 10 + 4);
  v.push_back(num * 10 + 7);
}
long long k, pl, pr, vl, vr;
inline long long get(long long l, long long r, long long a, long long b) {
  if (l > b || a > r) return 0;
  if (l >= a && b >= r) {
    return r - l + 1;
  }
  if (a >= l && b >= r) {
    return r - a + 1;
  }
  if (l >= a && r >= b) {
    return b - l + 1;
  }
  if (a >= l && r >= b) {
    return b - a + 1;
  }
  return 0;
}
void solve() {
  dfs(0, 0);
  v.push_back(0);
  v.push_back(10000000000);
  sort(v.begin(), v.end());
  int i = 1;
  int j = k;
  long long p = 0;
  while ((int)v.size() > j + 1) {
    p += (get(vl, vr, v[i - 1] + 1, v[i]) * get(pl, pr, v[j], v[j + 1] - 1));
    p += (get(pl, pr, v[i - 1] + 1, v[i]) * get(vl, vr, v[j], v[j + 1] - 1));
    if (k == 1 && v[i] >= vl && vr >= v[i] && v[i] >= pl && pr >= v[i]) p--;
    j++;
    i++;
  }
  printf("%.20Lf\n", p / (long double)((vr - vl + 1) * (pr - pl + 1)));
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  solve();
  return 0;
}