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, tot, a[1050];
void dfs(int num) {
  if (num) a[++tot] = num;
  if (num * 10ll + 4 <= max(pr, vr)) dfs(num * 10 + 4);
  if (num * 10ll + 7 <= max(pr, vr)) dfs(num * 10 + 7);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0);
  sort(a + 1, a + 1 + tot);
  if (k > tot) return puts("0"), 0;
  int64_t sum = (pr - pl + 1) * (vr - vl + 1ll), now = 0;
  auto gao = [](int a, int b, int c, int d) {
    int l = max(a, c), r = min(b, d);
    if (l <= r) return r - l + 1;
    return 0;
  };
  for (int l = 1, r = k; r <= tot; ++l, ++r) {
    int l1 = l == 1 ? 1 : a[l - 1] + 1, r1 = a[l];
    int l2 = a[r], r2 = r == tot ? max(pr, vr) : a[r + 1] - 1;
    now += gao(l1, r1, pl, pr) * (int64_t)gao(l2, r2, vl, vr);
    now += gao(l1, r1, vl, vr) * (int64_t)gao(l2, r2, pl, pr);
    if (k == 1 && pl <= a[l] && a[l] <= pr && vl <= a[l] && a[l] <= vr) --now;
  }
  printf("%.15lf\n", now / (double)sum);
}