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 a[3010];
long long cnt, pl, pr, vl, vr, k;
void dfs(int p, long long v) {
  if (p == 10)
    a[cnt++] = v;
  else {
    if (v) a[cnt++] = v;
    dfs(p + 1, v * 10 + 4);
    dfs(p + 1, v * 10 + 7);
  }
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  cnt = 1;
  dfs(0, 0);
  sort(a, a + cnt);
  double ans = 0;
  long long p = pr - pl + 1, v = vr - vl + 1;
  a[cnt] = (long long)1e18;
  for (int i = 1, j; i < cnt; i++)
    if (a[i] <= max(pr, vr) && (j = i + k - 1) < cnt && a[j] <= max(pr, vr)) {
      if (pl <= a[i] && pr >= a[i - 1]) {
        long long lef = max(pl, a[i - 1] + 1), rig = min((long long)pr, a[i]);
        long long ln = max(0ll, (rig - lef) + 1);
        long long rn =
            max(0ll, (min((long long)vr, a[j + 1] - 1) - max(vl, a[j])) + 1);
        ans += (double)ln / p * (double)rn / v;
      }
      if (vl <= a[i] && vr >= a[i - 1]) {
        long long ln =
            max(0ll, (min((long long)vr, a[i]) - max(vl, a[i - 1] + 1)) + 1);
        long long rn =
            max(0ll, (min((long long)pr, a[j + 1] - 1) - max(pl, a[j])) + 1);
        long long lef = max(pl, a[j]), rig = min((long long)pr, a[j + 1] - 1);
        ans += (double)ln / v * (double)rn / p;
      }
      if (k == 1 && a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr) {
        ans -= 1.0 / p * 1.0 / v;
      }
    }
  printf("%.10f\n", ans);
  return 0;
}