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;
void solve();
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
long long p1, p2, v1, v2, k;
long long happy[5000];
long long t = 0;
long long pere(pair<long long, long long> e, pair<long long, long long> r) {
  if (e > r) swap(e, r);
  if (e.second < r.first) return 0;
  return min(r.second, e.second) - max(r.first, e.first) + 1;
}
vector<pair<pair<long long, long long>, long long> > q, w;
void solve() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  long long p = p2 - p1 + 1;
  long long v = v2 - v1 + 1;
  happy[t++] = 0;
  for (long long i = 1; i <= 9; i++) {
    for (long long mask = 0; mask < (1 << i); mask++) {
      long long newnum = 0;
      for (long long j = 0; j < i; j++)
        if (mask & (1 << j))
          newnum = newnum * 10 + 7;
        else
          newnum = newnum * 10 + 4;
      happy[t++] = newnum;
    }
  }
  happy[t++] = 1000000001;
  sort(happy, happy + t);
  long long ind = 0;
  long long ans = 0;
  for (long long i = 0; i < t - 1; i++)
    if (happy[i] < p1 && happy[i + 1] >= p1) {
      ind = i;
      break;
    }
  for (long long i = ind; i + k < t - 1 && happy[i] <= p2; i++) {
    long long cur =
        pere(make_pair(happy[i + k], happy[i + k + 1] - 1), make_pair(v1, v2));
    long long z =
        pere(make_pair(happy[i] + 1, happy[i + 1]), make_pair(p1, p2));
    ans += z * cur;
  }
  ind = 0;
  for (long long i = 0; i < t - 1; i++)
    if (happy[i] < v1 && happy[i + 1] >= v1) {
      ind = i;
      break;
    }
  for (long long i = ind; i + k < t - 1 && happy[i] <= v2; i++) {
    long long cur =
        pere(make_pair(happy[i + k], happy[i + k + 1] - 1), make_pair(p1, p2));
    long long z =
        pere(make_pair(happy[i] + 1, happy[i + 1]), make_pair(v1, v2));
    ans += z * cur;
  }
  if (k == 1) {
    for (long long i = 0; i < t; i++) {
      long long cur = happy[i];
      if (cur >= p1 && cur <= p2 && cur <= v2 && cur >= v1) ans--;
    }
  }
  printf("%.10Lf", (long double)ans / p / v);
}