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 pl, pr, vl, vr, li, lj, ri, rj, re1, re2, ans = 0;
long long int vals[200000];
int k, t1 = 0;
long long int ret(long long int l, long long int l1, long long int r,
                  long long int r1) {
  l = max(l, l1);
  r = min(r, r1);
  if (r < l) {
    return 0;
  } else {
    return r - l + 1;
  }
}
int main() {
  vals[0] = 0;
  for (int i = 0; t1 < 200000; i++) {
    if (vals[i] * 10 + 4 > 1000000001) {
      t1++;
      vals[t1] = 1000000001;
      break;
    } else {
      t1++;
      vals[t1] = vals[i] * 10 + 4;
      t1++;
      vals[t1] = vals[i] * 10 + 7;
    }
  }
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 0; i < t1 - k; i++) {
    li = vals[i] + 1;
    lj = vals[i + 1];
    ri = vals[i + k];
    rj = vals[i + 1 + k] - 1;
    re1 = ret(li, pl, lj, pr) * ret(ri, vl, rj, vr);
    re2 = ret(ri, pl, rj, pr) * ret(li, vl, lj, vr);
    ans = ans + re1 + re2;
    if (re1 && re2 && k == 1) {
      ans--;
    }
  }
  printf("%.12lf\n", (double)ans / (vr - vl + 1ll) / (pr - pl + 1ll));
}