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 double pi = acos(-1.0);
const double eps = 1e-9;
const int inf = ~(1 << 31);
const int N = 1000000;
int luck[N], top = 1;
char s[16];
void gen(int i) {
  s[i] = '4';
  if (i == 0) {
    stringstream ss;
    ss << s;
    int k;
    ss >> k;
    luck[top++] = k;
  } else
    gen(i - 1);
  s[i] = '7';
  if (i == 0) {
    stringstream ss;
    ss << s;
    int k;
    ss >> k;
    luck[top++] = k;
  } else
    gen(i - 1);
  s[i] = 0;
}
double foo(double a, double b, double c, double d) {
  if (c <= a && d >= b) return b - a + 1;
  if (c >= a && d <= b) return d - c + 1;
  if (d >= a && c <= a) return d - a + 1;
  if (d >= b && c <= b) return b - c + 1;
  return 0;
}
int main() {
  int i, n, t;
  luck[0] = 0;
  for (i = 0; i < 9; i++) gen(i);
  luck[top] = inf;
  sort(luck, luck + top);
  double ul, ur, vl, vr;
  int k;
  cin >> ul >> ur >> vl >> vr >> k;
  double r = 0;
  for (i = 1; i + k - 1 < top; i++) {
    int j = i + k - 1;
    r += foo(luck[i - 1] + 1, luck[i], ul, ur) *
         foo(luck[j], luck[j + 1] - 1, vl, vr);
    r += foo(luck[i - 1] + 1, luck[i], vl, vr) *
         foo(luck[j], luck[j + 1] - 1, ul, ur);
  }
  if (k == 1) {
    for (i = 1; i < top; i++) {
      k = luck[i];
      if (k >= vl && k <= vr && k >= ul && k <= ur) r--;
    }
  }
  r /= double(vr - vl + 1);
  r /= double(ur - ul + 1);
  if (r > 1 + eps) r = 1.0;
  printf("%.9lf", r);
  return 0;
}