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 gen(long long num, int pl, int pr, int vl, int vr) {
  if (num > pr && num > vr) return;
  if (num >= min(pl, vl)) lucky.push_back(num);
  gen(num * 10 + 4, pl, pr, vl, vr);
  gen(num * 10 + 7, pl, pr, vl, vr);
}
long long go(int pos, int k, int vl, int vr, int pl, int pr) {
  long long count = 0;
  int leftleft, leftright, rightleft, rightright;
  if (lucky[pos] < vl || lucky[pos + k - 1] > pr) return 0LL;
  leftleft = vl;
  leftright = lucky[pos];
  rightleft = lucky[pos + k - 1];
  rightright = pr;
  if (pos > 0 && lucky[pos - 1] >= vl) leftleft = lucky[pos - 1] + 1;
  if (pos + k < (int)(lucky).size() && lucky[pos + k] <= pr)
    rightright = lucky[pos + k] - 1;
  if (vr < leftright) leftright = vr;
  if (pl > rightleft) rightleft = pl;
  if (leftleft >= vl && leftleft <= vr && leftright >= vl && leftright <= vr &&
      leftleft <= leftright && rightleft >= pl && rightleft <= pr &&
      rightright >= pl && rightright <= pr && rightleft <= rightright)
    count += (long long)(leftright - leftleft + 1) *
             (long long)(rightright - rightleft + 1);
  return count;
}
int main() {
  int pl, pr, vl, vr, k;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  gen(0, pl, pr, vl, vr);
  sort(lucky.begin(), lucky.end());
  long long count = 0;
  long long total = (long long)(pr - pl + 1) * (long long)(vr - vl + 1);
  if ((int)(lucky).size() < k)
    printf("0.0\n");
  else {
    int pos = 0;
    while (pos + k <= (int)(lucky).size()) {
      count += go(pos, k, vl, vr, pl, pr);
      count += go(pos, k, pl, pr, vl, vr);
      if (k == 1 && lucky[pos] >= vl && lucky[pos] <= vr && lucky[pos] >= pl &&
          lucky[pos] <= pr)
        count--;
      pos++;
    }
    printf("%.12f\n", (double)count / (double)(total));
  }
  return 0;
}