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 pl, pr, vl, vr, k;
int tot = 0, a[10001];
void pre() {
  for (int i = 1; i <= 9; i++)
    for (int j = 0; j < (1 << i); j++) {
      int now = 0;
      for (int k = i - 1; k >= 0; k--)
        if (j & (1 << k))
          now = now * 10 + 7;
        else
          now = now * 10 + 4;
      a[++tot] = now;
    }
  a[0] = 0;
  a[tot + 1] = 1e9;
}
long long calc(int x, int y, int l, int r) {
  int xx = max(x, l), yy = min(y, r);
  return max(0, yy - xx + 1);
}
int main() {
  pre();
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  long long ans = 0;
  for (int i = 1; i + k - 1 <= tot; i++) {
    int k1 = a[i - 1], k2 = a[i], k3 = a[i + k - 1], k4 = a[i + k];
    long long ans1 = calc(pl, pr, k1 + 1, k2), ans2 = calc(vl, vr, k3, k4 - 1);
    ans1 *= ans2;
    ans += ans1;
    ans1 = calc(vl, vr, k1 + 1, k2), ans2 = calc(pl, pr, k3, k4 - 1);
    ans1 *= ans2;
    ans += ans1;
    if (k == 1 && pl <= k2 && pr >= k2 && vl <= k2 && vr >= k2) ans--;
  }
  printf("%.10lf\n", ans * 1.0 / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}