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;
vector<int> v;
void gao(int dep, int val) {
  if (val != 0) v.push_back(val);
  if (dep == 9) return;
  gao(dep + 1, val * 10 + 4);
  gao(dep + 1, val * 10 + 7);
}
int pl, pr, vl, vr, k;
double gao() {
  double add = 0.0, ct = 0.0;
  int i;
  for (i = 0; i < v.size(); ++i) {
    int VL, VR;
    VL = v[i];
    VR = (i == v.size() - 1 ? 1000000005 : v[i + 1] - 1);
    VR = min(VR, pr);
    VL = max(VL, pl);
    if (VL > VR) continue;
    add = VR - VL + 1;
    if (i - k + 1 >= 0) {
      int nowl, nowr;
      if (i - k < 0)
        nowl = 1;
      else
        nowl = v[i - k] + 1;
      nowr = v[i - k + 1];
      int L = max(nowl, vl), R = min(nowr, vr);
      if (L <= R) {
        ct += add * (R - L + 1.0);
        if (k == 1) {
          if (v[i] >= VL && v[i] <= VR && v[i] >= L && v[i] <= R) ct -= 0.5;
        }
      }
    }
  }
  return ct;
}
int main() {
  gao(0, 0);
  sort(v.begin(), v.end());
  while (cin >> pl >> pr >> vl >> vr >> k) {
    int i;
    double ct = 0.0;
    ct += gao();
    swap(pl, vl);
    swap(pr, vr);
    ct += gao();
    printf("%.10f\n", ct / (((double)vr - vl + 1.0) * ((double)pr - pl + 1.0)));
  }
  return 0;
}