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 v[2][2], k;
long long cnt[2000], a[15], cnt1;
long long min(long long a, long long b) { return a > b ? b : a; }
void dfs(int i, int t) {
  if (i == t) {
    long long sum = 0;
    for (int j = 0; j < t; j++) {
      sum = (sum * 10) + a[j];
    }
    if (sum) cnt[cnt1++] = sum;
    return;
  } else {
    a[i] = 4;
    dfs(i + 1, t);
    a[i] = 7;
    dfs(i + 1, t);
  }
}
int find(int t, int num) {
  if (v[t][1] < cnt[num]) {
    return 0;
  } else {
    if (v[t][0] > cnt[num]) {
      if (num + 1 < cnt1) {
        if (cnt[num + 1] < v[t][0])
          return 0;
        else {
          if (v[t][1] >= cnt[num + 1]) {
            return cnt[num + 1] - v[t][0];
          } else {
            return v[t][1] + 1 - v[t][0];
          }
        }
      } else {
        return v[t][1] - v[t][0] + 1;
      }
    } else {
      if (num + 1 < cnt1) {
        if (v[t][1] >= cnt[num + 1]) {
          return cnt[num + 1] - cnt[num];
        } else {
          return v[t][1] + 1 - cnt[num];
        }
      } else {
        return v[t][1] - cnt[num] + 1;
      }
    }
  }
}
double work() {
  int i = 0;
  while (v[0][0] > cnt[i]) {
    i++;
  }
  double sum = 0;
  long long p1 = v[0][0];
  long long d = (v[1][1] - v[1][0] + 1) * (v[0][1] - v[0][0] + 1);
  for (; i <= cnt1 - k && v[0][1] >= p1; i++) {
    long long tmp = (min(cnt[i], v[0][1]) - p1 + 1) * find(1, i + k - 1);
    sum += double(tmp) / d;
    p1 = cnt[i] + 1;
  }
  i = 0;
  while (v[1][0] > cnt[i]) {
    i++;
  }
  p1 = v[1][0];
  for (; i <= cnt1 - k && v[1][1] >= p1; i++) {
    long long tmp = (min(cnt[i], v[1][1]) - p1 + 1) * find(0, i + k - 1);
    sum += double(tmp) / d;
    p1 = cnt[i] + 1;
  }
  if (k == 1) {
    for (int i = 0; i < cnt1; i++) {
      if (cnt[i] >= v[0][0] && cnt[i] <= v[0][1] && cnt[i] >= v[1][0] &&
          cnt[i] <= v[1][1]) {
        sum -= 1.0 / d;
      }
    }
  }
  return sum;
}
int main() {
  cnt1 = 0;
  while (scanf("%lld%lld%lld%lld%lld", &v[0][0], &v[0][1], &v[1][0], &v[1][1],
               &k) != EOF) {
    for (int i = 1; i < 10; i++) {
      dfs(0, i);
    }
    sort(cnt, cnt + cnt1);
    printf("%.9lf\n", work());
  }
  return 0;
}