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 int oo = 0x3f3f3f3f;
const double eps = 1e-9;
int nums[100100], M;
void gen(int v) {
  if (v) nums[M++] = v;
  if (v < 100000000) {
    gen(v * 10 + 4);
    gen(v * 10 + 7);
  }
}
int lo[2], hi[2], K;
int main() {
  M = 0;
  gen(0);
  nums[M++] = 0;
  nums[M++] = 1000000000;
  nums[M++] = 1000000000;
  sort(nums, nums + M);
  for (int i = (0); i < (2); i++) cin >> lo[i] >> hi[i];
  cin >> K;
  for (int i = (0); i < (2); i++) hi[i]++;
  long long res = 0;
  for (int kk = (0); kk < (2); kk++) {
    for (int n = (1); n < (M); n++) {
      long long a1 = min(hi[0], nums[n] + 1) - max(lo[0], nums[n - 1] + 1);
      if (a1 <= 0) continue;
      int to = n + K;
      if (to >= M) break;
      long long a2 = min(hi[1], nums[to]) - max(lo[1], nums[to - 1]);
      if (a2 <= 0) continue;
      res += a1 * a2;
    }
    swap(lo[0], lo[1]);
    swap(hi[0], hi[1]);
  }
  if (K == 1) {
    for (int i = (1); i < (M); i++)
      if (nums[i] >= lo[0] && nums[i] >= lo[1] && nums[i] < hi[0] &&
          nums[i] < hi[1]) {
        --res;
      }
  }
  long long div = 1;
  for (int i = (0); i < (2); i++) div *= hi[i] - lo[i];
  double erg = res / ((double)(div));
  printf("%.9lf\n", erg);
  return 0;
}