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 N = 1050;
long long p1, p2, v1, v2, x, sz, Num, R1S, R1E, R2S, R2E, R3S, R3E, R4S, R4E,
    arr[N];
double All, Good;
bool Valid1, Valid2, Valid3, Valid4, Same;
void Go(int length) {
  if (length > 9) return;
  arr[sz++] = Num;
  Num *= 10;
  Num += 4, Go(length + 1);
  Num += 3, Go(length + 1);
  Num /= 10;
}
void Prepare_Lucky_Numbers() {
  Go(0);
  arr[sz++] = 1000000000;
  sort(arr, arr + sz);
}
int main() {
  Prepare_Lucky_Numbers();
  scanf("%I64d %I64d %I64d %I64d %I64d", &p1, &p2, &v1, &v2, &x);
  All = (p2 - p1 + 1) * (v2 - v1 + 1);
  for (int i = 1; i < sz - x; i++) {
    long long St1 = arr[i - 1] + 1, En1 = arr[i], St2 = arr[i + x - 1],
              En2 = arr[i + x] - 1, tmp = 0;
    R1S = max(p1, St1), R1E = min(En1, p2), Valid1 = (R1S <= R1E);
    R2S = max(v1, St1), R2E = min(En1, v2), Valid2 = (R2S <= R2E);
    R3S = max(p1, St2), R3E = min(En2, p2), Valid3 = (R3S <= R3E);
    R4S = max(v1, St2), R4E = min(En2, v2), Valid4 = (R4S <= R4E);
    if (Valid1 && Valid4) {
      tmp += (R1E - R1S + 1) * (R4E - R4S + 1);
    }
    if (Valid2 && Valid3) {
      tmp += (R2E - R2S + 1) * (R3E - R3S + 1);
    }
    if (R2E == R3S && R1E == R4S && R1E == R2E) tmp--;
    Good += tmp;
  }
  printf("%.12lf", Good / All);
  return 0;
}