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;
int main() {
  set<int> S;
  for (int i = 0; i < (1 << 10); i++) {
    int x = 0;
    for (int j = 0; j < 9; j++) {
      x *= 10;
      if (i & (1 << j))
        x += 4;
      else
        x += 7;
    }
    do S.insert(x);
    while (x /= 10);
  }
  vector<int> V;
  V.push_back(0);
  for (set<int>::iterator it = S.begin(); it != S.end(); it++) V.push_back(*it);
  V.push_back(INT_MAX);
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double cnt = 0;
  double all = ((vr - vl) + 1.0) * ((pr - pl) + 1.0);
  for (int i = 1; i < V.size() - k; i++) {
    int hll = V[i - 1] + 1;
    int hl = V[i];
    int hr = V[(i + k) - 1];
    int hrr = V[(i + k)] - 1;
    if (pl <= hl && pr >= hll && vl <= hrr && vr >= hr) {
      cnt += (min(pr, hl) - max(pl, hll) + 1.0) *
             (min(vr, hrr) - max(vl, hr) + 1.0);
    }
    if (vl <= hl && vr >= hll && pl <= hrr && pr >= hr) {
      cnt += (min(vr, hl) - max(vl, hll) + 1.0) *
             (min(pr, hrr) - max(pl, hr) + 1.0);
      if (k == 1) cnt -= 1.0;
    }
  }
  double res = cnt / all;
  printf("%.12f", cnt / all);
exit:
  return (0);
}