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;
int q[1000100];
struct In {
  int l, r;
} p, v, a, b;
int gao(In x, In y) {
  In ret;
  ret.l = max(x.l, y.l);
  ret.r = min(x.r, y.r);
  return max(0, ret.r - ret.l + 1);
}
int i, j, k, K;
int main() {
  while (~scanf("%d%d%d%d%d", &p.l, &p.r, &v.l, &v.r, &K)) {
    int ct = 0;
    q[ct++] = min(p.l, v.l) - 1;
    for (i = 0; i < 10; i++) {
      for (j = 0; j < (1 << i); j++) {
        int te = 0;
        for (k = 0; k < i; k++) {
          te *= 10;
          if (j & (1 << k))
            te += 4;
          else
            te += 7;
        }
        if (te >= min(p.l, v.l) && te <= max(v.r, p.r)) q[ct++] = te;
      }
    }
    if (ct <= K) {
      puts("0");
      continue;
    }
    q[ct++] = max(p.r, v.r) + 1;
    sort(q, q + ct);
    long long ans = 0;
    for (i = 0; i + K + 1 < ct; i++) {
      a.l = q[i];
      if (i != 0) a.l++;
      a.r = q[i + 1];
      b.l = q[i + K];
      b.r = q[i + K + 1];
      if (i + K + 1 != ct - 1) b.r--;
      ans += (long long)gao(p, a) * gao(v, b);
      ans += (long long)gao(p, b) * gao(v, a);
      if (K == 1 && a.r >= p.l && a.r <= p.r && a.r >= v.l && a.r <= v.r)
        ans -= 1;
    }
    printf("%.11f\n",
           (double)ans / (double)(v.r - v.l + 1) / (double)(p.r - p.l + 1));
  }
}