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 long long N = 2e6 + 5;
const long long M = 5005;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x);
    x = (x * x);
    y >>= 1;
  }
  return ret;
}
long long luck[N], n;
void init() {
  for (long long L = 1; L <= 9; L++) {
    for (long long s = 0; s < (1LL << L); s++) {
      long long tmp = s, TMP = 0;
      for (long long i = 0; i < L; i++) {
        TMP += (4 + (tmp & 1) * 3) * ksm(10, i);
        tmp >>= 1;
      }
      luck[++n] = TMP;
    }
  }
}
long long pl, pr, vl, vr, k;
double ans;
long long calc(long long l, long long r, long long L, long long R) {
  if (r < L || l > R) return 0;
  if (l > r || L > R) return 0;
  if (l >= L && l <= R) return min(r, R) - l + 1;
  if (r >= L && r <= R) return r - max(l, L) + 1;
  if (l < L && r > R) return R - L + 1;
}
signed main() {
  init();
  luck[++n] = 0x3f3f3f3f3f3f3f3f;
  sort(luck + 1, luck + n + 1);
  cin >> pl >> pr >> vl >> vr >> k;
  for (long long i = 1; i + k - 1 <= n; i++) {
    if (luck[i] > vr && luck[i] > pr) break;
    long long j = i + k - 1;
    ans += calc(luck[i - 1] + 1, luck[i], pl, pr) *
           calc(luck[j], luck[j + 1] - 1, vl, vr);
    ans += calc(luck[j], luck[j + 1] - 1, pl, pr) *
           calc(luck[i - 1] + 1, luck[i], vl, vr);
    if (k == 1)
      if (calc(pl, pr, luck[i], luck[i]) && calc(vl, vr, luck[i], luck[i]))
        ans -= 1;
  }
  ans /= (pr - pl + 1);
  ans /= (vr - vl + 1);
  printf("%.12lf\n", ans);
}