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>
int lt[10], num[2000], dp[2000], dv[2000], mp[2000], mv[2000];
int n;
void genLuckyNum() {
  num[0] = 0;
  n = 1;
  lt[0] = 1;
  for (int i = 1; i < 9; i++) lt[i] = lt[i - 1] * 2;
  for (int i = 1; i <= 9; i++) {
    int en = (1 << i);
    for (int j = 0; j < en; j++) {
      int tmp = 0;
      for (int z = i - 1; z >= 0; z--) {
        tmp = tmp * 10 + ((j & lt[z]) ? 7 : 4);
      }
      num[n++] = tmp;
    }
  }
  num[n] = 1000000001;
}
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int main() {
  int pl, pr, vl, vr, k;
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  genLuckyNum();
  int vtp = 1;
  long long res = 0;
  while (vtp < n && pl > num[vtp]) vtp++;
  int vtv = 1;
  while (vtv < n && vl > num[vtv]) vtv++;
  {
    int pp, vv;
    pp = vtp;
    for (; pp <= n; pp++) {
      if (num[pp - 1] >= pr) break;
      int tmp1 = min(num[pp], pr) - max(num[pp - 1] + 1, pl) + 1;
      if (pp + k - 1 >= n || num[pp + k - 1] > vr) break;
      int tmp2 = min(num[pp + k] - 1, vr) - max(num[pp + k - 1], vl) + 1;
      if (num[pp + k] > vl) res += (long long)tmp1 * tmp2;
    }
    vv = vtv;
    for (; vv <= n; vv++) {
      if (num[vv - 1] >= vr) break;
      int tmp1 = min(num[vv], vr) - max(num[vv - 1] + 1, vl) + 1;
      if (vv + k - 1 >= n || num[vv + k - 1] > pr) break;
      int tmp2 = min(num[vv + k] - 1, pr) - max(num[vv + k - 1], pl) + 1;
      if (num[vv + k] > pl) res += (long long)tmp1 * tmp2;
    }
    if (k == 1) {
      for (int i = vtp; i < n; i++) {
        if (num[i] <= pr && num[i] >= vl && num[i] <= vr) res--;
      }
    }
  }
  double total = 1.0;
  total *= (pr - pl + 1);
  total *= (vr - vl + 1);
  printf("%.15lf", (res / total));
  return 0;
}