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> V;
void gen(int dig) {
  for (int i = 1; i <= dig; i++) {
    for (int j = 0; j < 1 << i; j++) {
      int num = 0;
      for (int k = 0; k < i; k++) {
        num *= 10;
        num += (j & 1 << k) ? 7 : 4;
      }
      V.push_back(num);
    }
  }
  sort(V.begin(), V.end());
}
int L0, L1, H0, H1, K;
int main() {
  gen(9);
  cin >> L0 >> H0 >> L1 >> H1 >> K;
  H0++;
  H1++;
  long long result = 0;
  for (int ii = 0; ii < 2; ii++) {
    for (int i = K - 1; i < V.size(); i++) {
      if (V[i] > H0) break;
      int LR0 = max(L0, V[i]);
      int HR0 = min(H0, i + 1 < V.size() ? V[i + 1] : 1000000010);
      int LR1 = max(L1, i - K >= 0 ? V[i - K] + 1 : 0);
      int HR1 = min(H1, V[i - K + 1] + 1);
      if (HR0 < LR0 || HR1 < LR1) continue;
      result += 1ll * (HR0 - LR0) * (HR1 - LR1);
    }
    swap(L0, L1);
    swap(H0, H1);
  }
  if (K == 1) {
    for (int i = 0; i < V.size(); i++) {
      if (L0 <= V[i] && V[i] < H0 && L1 <= V[i] && V[i] < H1) {
        result--;
      }
    }
  }
  printf("%.12lf\n", 1.0 * result / (H0 - L0) / (H1 - L1));
  return 0;
}