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;
long long int gcd(long long int a, long long int b) {
  if (b < a) {
    long long int temp = a;
    a = b;
    b = temp;
  }
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int lcm(long long int a, long long int b) {
  return (a * b / gcd(a, b));
}
long long int fact(long long int x) {
  long long int ans = 1;
  if (x == 0) {
    return 1;
  }
  while (x > 0) {
    ans = (ans * x) % 1000000007;
    x--;
  }
  return ans % 1000000007;
}
long long int comb(long long int n, long long int r) {
  long long int x = fact(r), y = fact(n - r);
  x = (x * y) % 1000000007;
  if (fact(n) > x) return fact(n) / x;
  return (fact(n) + 1000000007) / x;
}
long long int power(long long int a, long long int p) {
  long long int ans = 1;
  while (p > 0) {
    if (p % 2) {
      ans *= a;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    p /= 2;
  }
  return ans % 1000000007;
}
float dist(long long int x1, long long int y1, long long int x2,
           long long int y2) {
  return sqrt(1.0 * (x1 - x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2));
}
vector<long long int> getbinary(long long int x) {
  vector<long long int> bin(20000, 0);
  long long int iter = 0;
  while (x > 0) {
    if (x % 2 == 1) {
      bin[iter] = 1;
    } else {
      bin[iter] = 0;
    }
    iter++;
    x /= 2;
  }
  return bin;
}
long long int fn(long long int l1, long long int r1, long long int l2,
                 long long int r2) {
  l1 = max(l1, l2);
  r1 = min(r1, r2);
  return max(r1 - l1 + 1, 0ll);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int l1, r1, l2, r2, k = 0, a[5000], i, mi, ma, d = 0;
  a[0] = 0;
  for (i = 1; i < 5000; i += 2) {
    a[i] = a[k] * 10 + 4;
    a[i + 1] = a[k] * 10 + 7;
    k++;
  }
  cin >> l1 >> r1 >> l2 >> r2;
  cin >> k;
  for (i = 1; i < 5000 - k; i++) {
    mi =
        fn(a[i - 1] + 1, a[i], l2, r2) * fn(a[i + k - 1], a[i + k] - 1, l1, r1);
    ma =
        fn(a[i - 1] + 1, a[i], l1, r1) * fn(a[i + k - 1], a[i + k] - 1, l2, r2);
    d += ma + mi;
    if (k == 1 && mi > 0 && ma > 0) d--;
  }
  long long int all = (r1 - l1 + 1) * (r2 - l2 + 1);
  cout << fixed << setprecision(12) << 1.0 * d / all;
}