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 int max_n = 11, inf = 1000000000;
int pl, pr, vl, vr, k, L, R;
long long ans;
vector<int> v;
void gen(int x, int cnt) {
  v.push_back(x);
  if (cnt == 9) {
    return;
  }
  gen(x * 10 + 4, cnt + 1);
  gen(x * 10 + 7, cnt + 1);
}
int cnt(int l1, int r1, int l2, int r2) {
  if (l1 > l2) {
    swap(l1, l2);
    swap(r1, r2);
  }
  if (l2 > r1) {
    return 0;
  }
  return min(r1, r2) - l2 + 1;
}
void proc(int l1, int r1, int l2, int r2) {
  if (l1 > l2) {
    swap(l1, l2);
    swap(r1, r2);
  }
  if (l2 > r1) {
    return;
  }
  L = l2;
  R = min(r1, r2);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  proc(pl, pr, vl, vr);
  gen(0, 0);
  v.push_back(inf + 10);
  sort(v.begin(), v.end());
  for (int i = 1; i + k < v.size(); ++i) {
    int l1 = v[i - 1] + 1;
    int r1 = v[i];
    int l2 = v[i + k - 1];
    int r2 = v[i + k] - 1;
    long long c1 = cnt(l1, r1, pl, pr);
    long long c2 = cnt(l2, r2, vl, vr);
    ans += c1 * c2;
    c1 = cnt(l2, r2, pl, pr);
    c2 = cnt(l1, r1, vl, vr);
    ans += c1 * c2;
    if (k == 1 && L <= r1 && r1 <= R) ans -= 1;
  }
  printf("%.10f\n", 1.0 * ans / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}