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> happy;
void rec(int md, int d = 0, long long x = 0) {
  if (d == md) {
    happy.push_back(x);
    return;
  }
  for (int i = 0; i < 2; i++) {
    x *= 10;
    if (i)
      x += 7;
    else
      x += 4;
    rec(md, d + 1, x);
    x /= 10;
  }
  return;
}
long long pl, pr, vl, vr, k;
int main() {
  long long res = 0;
  happy.push_back(0);
  for (long long i = 1; i < 11; i++) rec(i);
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 1; i < happy.size() - k; i++) {
    long long a = (min(pr, happy[i]) - max(happy[i - 1] + 1, pl) + 1);
    long long b = (min(vr, happy[i + k] - 1) - max(happy[i + k - 1], vl) + 1);
    long long c = (min(vr, happy[i]) - max(happy[i - 1] + 1, vl) + 1);
    long long d = (min(pr, happy[i + k] - 1) - max(happy[i + k - 1], pl) + 1);
    if (a > 0 && b > 0) res += a * b;
    if (c > 0 && d > 0) res += c * d;
    if (k == 1 && pl <= happy[i] && happy[i] <= pr && vl <= happy[i] &&
        happy[i] <= vr)
      res--;
  }
  printf("%.12lf", (double)res / (double)((pr - pl + 1) * (vr - vl + 1)));
  return 0;
}