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;
using LL = long long;
const int M = 1e5 + 5;
const int N = 1e2 + 2;
const int mo = 998244353;
int a[M], cnt;
void init() {
  a[cnt++] = 0;
  for (int n = 1; n <= 9; n++) {
    int ls = (1 << n);
    for (int i = 0; i < ls; i++) {
      int res = 0;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j))
          res = res * 10 + 7;
        else
          res = res * 10 + 4;
      }
      a[cnt++] = res;
    }
  }
  a[cnt++] = 1e9 + 1;
  sort(a, a + cnt);
}
int main() {
  init();
  int pl, pr, vl, vr, k;
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  double len1 = pr - pl + 1, len2 = vr - vl + 1;
  double res = 0;
  for (int i = 0; i + k < cnt; i++) {
    if (a[i + 1] >= pl && pr > a[i]) {
      if (a[i + k + 1] > vl && vr >= a[i + k]) {
        double low1 = max(pl, a[i] + 1), up1 = min(a[i + 1], pr);
        double low2 = max(a[i + k], vl), up2 = min(a[i + k + 1] - 1, vr);
        res += ((up1 - low1 + 1) / len1) * ((up2 - low2 + 1) / len2);
      }
    }
    if (a[i + 1] >= vl && vr > a[i]) {
      if (a[i + k + 1] > pl && pr >= a[i + k]) {
        double low1 = max(vl, a[i] + 1), up1 = min(a[i + 1], vr);
        double low2 = max(a[i + k], pl), up2 = min(a[i + k + 1] - 1, pr);
        res += ((up1 - low1 + 1) / len2) * ((up2 - low2 + 1) / len1);
      }
    }
  }
  if (k == 1) {
    for (int i = 1; i + 1 < cnt; i++) {
      if (a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr)
        res -= (1.0 / len1) * (1.0 / len2);
    }
  }
  printf("%.10f\n", res);
  return 0;
}