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 pl, pr, vl, vr;
int k;
long long q[3000];
vector<long long> val;
void bfs() {
  int b = 0, e = 2;
  q[0] = 4ll, q[1] = 7ll;
  while (q[b] <= 1000000000) {
    val.push_back(q[b]);
    q[e++] = q[b] * 10 + 4;
    q[e++] = q[b] * 10 + 7;
    b++;
  }
}
int main(void) {
  scanf(" %I64d %I64d %I64d %I64d %d", &pl, &pr, &vl, &vr, &k);
  val.push_back(0);
  bfs();
  val.push_back(1010000000);
  long double ans = 0.0;
  int b = 1, e = k;
  long long ch = 0;
  int cnt1 = 0;
  int cnt2 = 0;
  long long kk = 0;
  if (pl >= vl && pr <= vr)
    kk = pr - pl + 1;
  else if (vl >= pl && vr <= pr)
    kk = vr - vl + 1;
  else
    kk = max(0ll, min(vr, pr) - max(vl, pl) + 1);
  while (e < ((int)(val.size()))) {
    ch +=
        max(min(val[b], (long long)pr) - max((long long)pl, val[b - 1] + 1) + 1,
            0ll) *
        max(0ll, min((long long)vr, val[e + 1] - 1) -
                     max(val[e], (long long)vl) + 1);
    ch +=
        max(min(val[b], (long long)vr) - max((long long)vl, val[b - 1] + 1) + 1,
            0ll) *
        max(0ll, min((long long)pr, val[e + 1] - 1) -
                     max(val[e], (long long)pl) + 1);
    ans += (double)(max(ch, 0ll)) / (double)((pr - pl + 1) * (vr - vl + 1));
    ch = 0;
    e++, b++;
  }
  kk = 0;
  for (int i = 1; i < ((int)(val.size())) - 1; i++) {
    if (k == 1 && val[i] >= vl && val[i] >= pl && val[i] <= pr && val[i] <= vr)
      kk++;
  }
  printf("%.18lf",
         (double)ans - (double)(kk) / (double)((pr - pl + 1) * (vr - vl + 1)));
  return 0;
}