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;
const long long INF = 1000000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
long long PL, PR, VL, VR, K;
vector<long long> V;
void proc(long long n) {
  if (n > INF * 10) return;
  if (n > 0) V.push_back(n);
  proc(n * 10 + 4);
  proc(n * 10 + 7);
}
int main() {
  proc(0);
  sort(V.begin(), V.end());
  cin >> PL >> PR >> VL >> VR >> K;
  double ans = 0;
  for (int b = (K - 1); b <= (((int)V.size()) - 2); b++) {
    long long l1 = b == K - 1 ? 1 : V[b - K] + 1;
    long long r1 = V[b - K + 1];
    long long l2 = V[b];
    long long r2 = V[b + 1] - 1;
    long long com_l = max(l1, PL);
    long long com_r = min(r1, PR);
    double nums = max((double)com_r - com_l + 1, 0.0);
    double p = nums / (PR - PL + 1);
    com_l = max(l2, VL);
    com_r = min(r2, VR);
    nums = max((double)com_r - com_l + 1, 0.0);
    p *= nums / (VR - VL + 1);
    ans += p;
    com_l = max(l1, VL);
    com_r = min(r1, VR);
    nums = max((double)com_r - com_l + 1, 0.0);
    p = nums / (VR - VL + 1);
    com_l = max(l2, PL);
    com_r = min(r2, PR);
    nums = max((double)com_r - com_l + 1, 0.0);
    p *= nums / (PR - PL + 1);
    ans += p;
    if (l2 == r1) {
      double p1 = r1 >= VL && r1 <= VR ? 1.0 / (VR - VL + 1) : 0;
      double p2 = r1 >= PL && r1 <= PR ? 1.0 / (PR - PL + 1) : 0;
      ans -= p1 * p2;
    }
  }
  cout << setprecision(15) << fixed << ans;
}