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;
long long lucky[2048];
int len;
int l1, r1, l2, r2, k;
long long ans;
int itv_and(long long l1, long long r1, long long l2, long long r2);
void init();
void gao();
void print();
void prepare();
long long calc(int l1, int r1, int l2, int r2);
int main() {
  prepare();
  init();
  gao();
  print();
  return 0;
}
void init() {
  scanf("%d%d%d%d%d", &l1, &r1, &l2, &r2, &k);
  if (l2 <= l1) swap(l1, l2), swap(r1, r2);
}
void print() { printf("%.10lf\n", 1.0 * ans / (r1 - l1 + 1) / (r2 - l2 + 1)); }
void prepare() {
  int i, j, s, t;
  lucky[0] = 0;
  len = 0;
  s = t = 0;
  for (i = 1; i <= 10; i++) {
    for (j = s; j <= t; j++) {
      lucky[++len] = lucky[j] * 10 + 4;
      lucky[++len] = lucky[j] * 10 + 7;
    }
    s = t + 1;
    t = len;
  }
}
int itv_and(long long l1, long long r1, long long l2, long long r2) {
  long long l = max(l1, l2);
  long long r = min(r1, r2);
  if (l > r) return 0;
  return r - l + 1;
}
void gao() {
  ans = 0;
  if (l2 > r1)
    ans = calc(l1, r1, l2, r2);
  else if (r1 > r2)
    ans = calc(l1, l2 - 1, l2, r2) + calc(l2, r2, l2, r2) +
          calc(l2, r2, r2 + 1, r1);
  else {
    ans = calc(l1, l2 - 1, l2, r2) + calc(l2, r1, l2, r1) +
          calc(l2, r1, r1 + 1, r2);
  }
}
long long calc(int l1, int r1, int l2, int r2) {
  if (l1 > r1 || l2 > r2) return 0;
  int i, p;
  long long ret, left, right, itv_l, itv_r;
  ret = 0;
  for (i = 1; i <= len; i++)
    if (lucky[i] >= l1) {
      p = i;
      break;
    }
  left = l1;
  do {
    if (p > len) break;
    right = lucky[p];
    if (right > r1) right = r1;
    if (p + k - 1 > len || lucky[p + k - 1] > r2) break;
    itv_l = lucky[p + k - 1];
    if (p + k > len) break;
    itv_r = lucky[p + k] - 1;
    if (itv_l > itv_r) break;
    ret += 1LL * (right - left + 1) * itv_and(l2, r2, itv_l, itv_r);
    left = right + 1;
    p++;
  } while (left <= r1);
  if (l1 == l2 && r1 == r2) {
    ret = ret + ret;
    if (k == 1)
      for (i = 1; i <= len; i++) {
        if (lucky[i] >= l1 && lucky[i] <= r1) ret--;
        if (lucky[i] > r1) break;
      }
  }
  return ret;
}