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;
vector<int> lucky;
const int INF = 1 << 30;
void init() {
  lucky.push_back(0);
  for (int l = 1; l <= 9; ++l)
    for (int b = 0; b < (1 << l); ++b) {
      int k = 0;
      for (int i = 0, f = 1; i < l; ++i, f *= 10) {
        if (b & 1 << i)
          k += f * 7;
        else
          k += f * 4;
      }
      lucky.push_back(k);
    }
  lucky.push_back(INF);
}
int main() {
  int p[2], v[2], k;
  init();
  scanf("%d%d%d%d%d", &p[0], &p[1], &v[0], &v[1], &k);
  long long sum = 0;
  int n = int(lucky.size()) - 1;
  for (int a = 1, b = k; b < n; ++a, ++b) {
    int x[2], y[2], d[2];
    x[0] = max(p[0], lucky[a - 1] + 1);
    x[1] = min(p[1], lucky[a]);
    y[0] = max(v[0], lucky[b]);
    y[1] = min(v[1], lucky[b + 1] - 1);
    d[0] = x[1] - x[0] + 1;
    d[1] = y[1] - y[0] + 1;
    if (d[0] > 0 && d[1] > 0) sum += (long long)d[0] * d[1];
    int s[2], t[2];
    s[0] = max(v[0], lucky[a - 1] + 1);
    s[1] = min(v[1], lucky[a]);
    t[0] = max(p[0], lucky[b]);
    t[1] = min(p[1], lucky[b + 1] - 1);
    d[0] = s[1] - s[0] + 1;
    d[1] = t[1] - t[0] + 1;
    if (d[0] > 0 && d[1] > 0) sum += (long long)d[0] * d[1];
    int xx[2], yy[2];
    xx[0] = max(x[0], t[0]);
    xx[1] = min(x[1], t[1]);
    yy[0] = max(y[0], s[0]);
    yy[1] = min(y[1], s[1]);
    d[0] = xx[1] - xx[0] + 1;
    d[1] = yy[1] - yy[0] + 1;
    if (d[0] > 0 && d[1] > 0) sum -= (long long)d[0] * d[1];
  }
  double ans = sum / (1.0 * (p[1] - p[0] + 1) * (v[1] - v[0] + 1));
  printf("%.10lf\n", ans);
}