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<long long> lucky;
void backtrack(long long num) {
  if (num > 10000000000) return;
  lucky.push_back(num);
  backtrack(num * 10 + 4);
  backtrack(num * 10 + 7);
}
int main() {
  backtrack(0);
  sort(lucky.begin(), lucky.end());
  pair<long long, long long> P, V;
  int k;
  cin >> P.first >> P.second;
  cin >> V.first >> V.second;
  cin >> k;
  long long denominator =
      1LL * (P.second - P.first + 1) * (V.second - V.first + 1);
  long long numerator = 0;
  for (int i = k; i + 1 < (int)lucky.size(); i++) {
    long long pLo = max(lucky[i - k] + 1, P.first);
    long long pHi = min(lucky[i - k + 1], P.second);
    long long vLo = max(lucky[i], V.first);
    long long vHi = min(lucky[i + 1] - 1, V.second);
    vLo = max(vLo, max(pLo, pHi));
    long long yo = 1LL * max(0LL, pHi - pLo + 1) * max(0LL, vHi - vLo + 1);
    numerator += yo;
    pLo = max(lucky[i], P.first);
    pHi = min(lucky[i + 1] - 1, P.second);
    vLo = max(lucky[i - k] + 1, V.first);
    vHi = min(lucky[i - k + 1], V.second);
    pLo = max(pLo, max(vLo, vHi));
    yo = 1LL * max(0LL, pHi - pLo + 1) * max(0LL, vHi - vLo + 1);
    numerator += yo;
  }
  if (k == 1)
    for (int i = 0; i < (int)lucky.size(); i++)
      if (lucky[i] >= max(P.first, V.first) &&
          lucky[i] <= min(P.second, V.second))
        numerator -= 1;
  printf("%.9lf\n", (1.0 * numerator) / denominator);
  return 0;
}