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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
long long A[(1 << 14)], T;
int L = 1, b, e, B, E, K;
void rec(long long S, int r) {
  if (!r--) {
    A[L++] = S;
    return;
  }
  rec(S * 10 + 4, r), rec(S * 10 + 7, r);
}
long long fs(int x, int y) { return max(0, min(y, E) - max(B, x) + 1); }
long long sc(int x, int y) { return max(0, min(y, e) - max(b, x) + 1); }
int main(void) {
  for (int k(1); k < 10; k++) rec(0, k);
  sort(A, A + L), A[L++] = int(1e9 + 1);
  scanf("%d%d%d%d%d", &b, &e, &B, &E, &K);
  for (int k(1); k < L - 1; k++)
    if (k + K < L) {
      T += fs(A[k - 1] + 1, A[k]) * sc(A[k + K - 1], A[k + K] - 1);
      T += sc(A[k - 1] + 1, A[k]) * fs(A[k + K - 1], A[k + K] - 1);
      if (K == 1) T -= fs(A[k], A[k]) * sc(A[k], A[k]);
    }
  printf("%.30lf\n", min(1., T / ((E - B + 1.) * (e - b + 1.))));
  return 0;
}