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 luck[100100];
int cnt;
long long low(int i, int j, long long pl, long long pr) {
  long long l = max(luck[i - 1] + 1, pl);
  long long r = min(luck[i], pr);
  if (l <= r) {
    return r - l + 1;
  }
  return 0;
}
long long high(int i, int j, long long pl, long long pr) {
  long long r = min(luck[j + 1] - 1, pr);
  long long l = max(pl, luck[j]);
  if (l <= r) {
    return r - l + 1;
  }
  return 0;
}
int num[2] = {4, 7};
int res[20];
int main() {
  long long pl, pr, vl, vr, m;
  cnt = 0;
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < 1 << i; j++) {
      vector<int> bit;
      int n = j;
      for (int k = 0; k < i; k++) {
        bit.push_back(n % 2);
        n /= 2;
      }
      luck[++cnt] = 0;
      for (int k = bit.size() - 1; k >= 0; k--) {
        luck[cnt] = luck[cnt] * 10 + (bit[k] ? 7 : 4);
      }
    }
  }
  luck[++cnt] = 1000000001;
  sort(luck, luck + cnt + 1);
  scanf("%I64d%I64d%I64d%I64d%I64d", &pl, &pr, &vl, &vr, &m);
  long long ans = 0;
  for (int i = 1; i < cnt - m + 1; i++) {
    long long l1 = low(i, i + m - 1, pl, pr), r1 = high(i, i + m - 1, vl, vr);
    ans += l1 * r1;
    long long l2 = low(i, i + m - 1, vl, vr), r2 = high(i, i + m - 1, pl, pr);
    ans += l2 * r2;
    if (m == 1) {
      if (pl <= luck[i] && pr >= luck[i] && vl <= luck[i] && vr >= luck[i]) {
        ans--;
      }
    }
  }
  printf("%.20lf\n", (double)ans / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}