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;
const int N = 2000;
int pl, pr, vl, vr, k, n;
long long a[N];
double res;
void Dfs(int u, int v, long long s) {
  if (u > v) {
    ++n;
    a[n] = s;
    return;
  }
  Dfs(u + 1, v, s * 10 + 4);
  Dfs(u + 1, v, s * 10 + 7);
}
inline long long Calc(long long x, long long y, long long xx, long long yy) {
  if (y < xx) return 0;
  if (yy < x) return 0;
  return min(y, yy) - max(x, xx) + 1;
}
int main() {
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  for (int i = 1; i <= 9; i++) Dfs(1, i, 0);
  sort(a + 1, a + 1 + n);
  res = 0;
  a[0] = 0;
  a[1023] = 2147483647;
  for (int i = 1; i + k - 1 <= n; i++) {
    res += (double)(Calc(a[i - 1] + 1, a[i], pl, pr)) / (double)(pr - pl + 1) *
           (double)(Calc(a[i + k - 1], a[i + k] - 1, vl, vr)) /
           (double)(vr - vl + 1);
    res += (double)(Calc(a[i - 1] + 1, a[i], vl, vr)) / (double)(vr - vl + 1) *
           (double)(Calc(a[i + k - 1], a[i + k] - 1, pl, pr)) /
           (double)(pr - pl + 1);
    if (k == 1 && a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr)
      res -= (double)(1) / (double)(pr - pl + 1) / (double)(vr - vl + 1);
  }
  printf("%.12lf\n", res);
  return 0;
}