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>
const int BUFFSIZE = 10240;
char BUFF[BUFFSIZE + 1], *p = BUFF;
int CHAR, SIGN, BYTES = 0;
using namespace std;
vector<long long> V;
void dfs(long long N) {
  if (N > 5000000000LL) return;
  V.push_back(N);
  dfs(N * 10 + 4);
  dfs(N * 10 + 7);
}
int main() {
  dfs(0);
  sort(V.begin(), V.end());
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double den = (pr - pl + 1) * (vr - vl + 1);
  double num = 0;
  for (int i = 1; i + k - 1 < V.size(); i++) {
    long long x = V[i];
    long long y = V[i + k - 1];
    long long lftp = V[i - 1] + 1;
    long long rgtp = x;
    long long lftv = y;
    long long rgtv = V[i + k] - 1;
    lftp = max(lftp, pl);
    rgtp = min(rgtp, pr);
    lftv = max(lftv, vl);
    rgtv = min(rgtv, vr);
    if (lftp <= rgtp && lftv <= rgtv) {
      double prob = (rgtp - lftp + 1.0) * (rgtv - lftv + 1.0);
      num += prob;
    }
    lftv = V[i - 1] + 1;
    rgtv = x;
    lftp = y;
    rgtp = V[i + k] - 1;
    lftp = max(lftp, pl);
    rgtp = min(rgtp, pr);
    lftv = max(lftv, vl);
    rgtv = min(rgtv, vr);
    if (lftp <= rgtp && lftv <= rgtv) {
      double prob = (rgtp - lftp + 1.0) * (rgtv - lftv + 1.0);
      num += prob;
    }
  }
  if (k == 1) {
    for (int i = 1; i < V.size(); i++)
      if (V[i] >= pl && V[i] <= pr && V[i] >= vl && V[i] <= vr) num -= 1;
  }
  double res = 1.0 * num / den;
  printf("%.20lf\n", res);
  return 0;
}