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 maxint = 0x7fffffff;
int m = 0, border = 1000000000, l1, r1, l2, r2, k;
long long g[500000];
void calc(long long v) {
  if (v > border) return;
  g[m++] = v;
  calc(v * 10 + 4);
  calc(v * 10 + 7);
}
long long work(long long t1, long long t2, long long t3, long long t4) {
  long long ll, rr;
  ll = max(t1, t3);
  rr = min(t2, t4);
  if (ll > rr) return 0;
  return rr - ll + 1;
}
int main() {
  calc(0);
  sort(g, g + m);
  g[m] = border + 1;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  long long ans = 0, vv;
  vv = r1 - l1 + 1;
  vv *= r2 - l2 + 1;
  for (int i = 1; i + k - 1 < m; i++) {
    long long t1, t3, t2, t4;
    t1 = g[i - 1] + 1;
    t2 = g[i];
    t3 = g[i + k - 1];
    t4 = g[i + k] - 1;
    long long k1, k2;
    k1 = work(t1, t2, l1, r1) * work(t3, t4, l2, r2);
    k2 = work(t1, t2, l2, r2) * work(t3, t4, l1, r1);
    ans += k1 + k2;
    if (k == 1 && k1 && k2) ans--;
  }
  printf("%.12f\n", ans * 1. / vv);
}