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 maxn = 1e5 + 5;
const int mod = 1e9 + 7;
long long p1, p2, v1, v2, l, r, x1, x2, x3, x4;
long long anss;
int k;
vector<long long> ans;
void dfs(long long x, int pos) {
  ans.push_back(x);
  if (pos >= 11) return;
  dfs(x * 10 + 4, pos + 1);
  dfs(x * 10 + 7, pos + 1);
}
long long sum(long long L, long long R, long long l, long long r) {
  if (r < L || l > R) return 0;
  long long x = max(l, L);
  long long y = min(r, R);
  return y - x + 1;
}
int main() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  dfs(0, 1);
  sort(ans.begin(), ans.end());
  int n = ans.size();
  for (int i = 0; i + k + 1 < n; i++) {
    l = ans[i] + 1, r = ans[i + 1];
    x1 = sum(p1, p2, l, r);
    x3 = sum(v1, v2, l, r);
    l = ans[i + k], r = ans[i + k + 1] - 1;
    x2 = sum(v1, v2, l, r);
    x4 = sum(p1, p2, l, r);
    anss += x1 * x2;
    anss += x3 * x4;
    if (k == 1 && sum(p1, p2, ans[i + 1], ans[i + 1]) &&
        sum(v1, v2, ans[i + 1], ans[i + 1]))
      anss--;
  }
  printf("%.13f\n", anss * 1.0 / (p2 - p1 + 1) / (v2 - v1 + 1));
  return 0;
}