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 pl, pr;
int vl, vr;
int k;
vector<int> v;
int T = 100000000;
void Fun(int x) {
  if (x != 0) v.push_back(x);
  if (x >= T) return;
  Fun(x * 10 + 4);
  Fun(x * 10 + 7);
}
double ans;
double Prob(int l, int r, int len) {
  if (l > r) return 0;
  return (r - l + 1) / (double)len;
}
void Solve(int l, int r, int bl, int br) {
  double res1, res2, res3;
  if (l >= pl && r <= vr) {
    res1 = Prob(max(pl, bl + 1), min(pr, l), pr - pl + 1) *
           Prob(max(vl, r), min(vr, br - 1), vr - vl + 1);
  } else {
    res1 = 0;
  }
  if (l >= vl && r <= pr) {
    res2 = Prob(max(vl, bl + 1), min(vr, l), vr - vl + 1) *
           Prob(max(pl, r), min(pr, br - 1), pr - pl + 1);
  } else {
    res2 = 0;
  }
  if (k == 1 && vl <= l && l <= vr && pl <= l && l <= pr) {
    res3 = Prob(l, l, pr - pl + 1) * Prob(l, l, vr - vl + 1);
  } else {
    res3 = 0;
  }
  ans += res1 + res2 - res3;
}
int main() {
  int i;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  T = min(T, max(pr, vr));
  Fun(0);
  v.push_back(0);
  v.push_back(1000000000);
  sort(v.begin(), v.end());
  ans = 0;
  for (i = 1; i <= (int)v.size() - k - 1; i++) {
    Solve(v[i], v[i + k - 1], v[i - 1], v[i + k]);
  }
  printf("%.12lf\n", ans);
  return 0;
}