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[20005];
long long k = 0;
void dfs(long long x) {
  if (x > 1000000000) return;
  a[++k] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
int main() {
  dfs(4);
  dfs(7);
  sort(a + 1, a + k + 1);
  a[k + 1] = 1000000000000;
  long long l, r, L, R, m;
  cin >> l >> r >> L >> R >> m;
  if (r > R) {
    swap(r, R);
    swap(l, L);
  }
  long long ans = 0;
  for (long long i = 1; i <= k - m + 1; i++) {
    if (a[i] >= l && a[i] <= r && a[i + m - 1] <= R && L < a[i + m]) {
      long long xx = max(L, a[i + m - 1]);
      ans = ans +
            min(a[i] - a[i - 1], a[i] - l + 1) * min(a[i + m] - xx, R - xx + 1);
    }
  }
  long long h = lower_bound(a + 1, a + k + 1, r) - a;
  if (h <= k && r != a[h] && h + m - 1 <= k) {
    long long xx = min(r - a[h - 1], r - l + 1);
    if (L < a[h + m] && R >= a[h + m - 1]) {
      long long xxx = max(L, a[h + m - 1]);
      ans = ans + xx * min(a[h + m] - xxx, R - xxx + 1);
    }
  }
  for (long long i = 1; i <= k - m + 1; i++) {
    if (a[i] >= L && a[i] <= R && a[i + m - 1] <= r && l < a[i + m]) {
      long long xx = max(l, a[i + m - 1]);
      ans = ans +
            min(a[i] - a[i - 1], a[i] - L + 1) * min(a[i + m] - xx, r - xx + 1);
    }
  }
  h = lower_bound(a + 1, a + k + 1, R) - a;
  if (h <= k && R != a[h] && h + m - 1 <= k) {
    long long xx = min(R - a[h - 1], R - L + 1);
    if (l < a[h + m] && r >= a[h + m - 1]) {
      long long xxx = max(L, a[h + m - 1]);
      ans = ans + xx * min(a[h + m] - xxx, r - xxx + 1);
    }
  }
  if (m == 1) {
    long long s = max(l, L);
    long long w = min(r, R);
    for (int i = 1; i <= k; i++) {
      if (a[i] >= s && a[i] <= w) ans--;
    }
  }
  double sum = (r - l + 1) * (R - L + 1);
  double x = ans;
  x /= sum;
  if (x > 1) x = 1;
  printf("%.9lf\n", x);
}