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;
typedef vector<vector<int>> graph;
ifstream in("buffcraft.in", ios::in);
ofstream out("buffcraft.out", ios::out);
int pl, pr, vl, vr, k;
vector<int> p;
void dfs(long long x) {
  if (x < 1000000000) {
    p.push_back(x);
    dfs(x * 10 + 4);
    dfs(x * 10 + 7);
  }
}
int main() {
  dfs(0);
  p.push_back(INT_MAX);
  sort(p.begin(), p.end());
  cin >> pl >> pr >> vl >> vr >> k;
  double cnt = 0;
  double all = ((vr - vl) + 1.0) * ((pr - pl) + 1.0);
  for (int i = 1; i < p.size() - k; i++) {
    int lm = p[i - 1] + 1, l = p[i], r = p[i + k - 1], rp = p[i + k] - 1;
    if (pl <= l && pr >= lm && vl <= rp && vr >= r)
      cnt +=
          (min(pr, l) - max(pl, lm) + 1.0) * (min(vr, rp) - max(vl, r) + 1.0);
    if (vl <= l && vr >= lm && pl <= rp && pr >= r) {
      cnt +=
          (min(vr, l) - max(vl, lm) + 1.0) * (min(pr, rp) - max(pl, r) + 1.0);
      if (k == 1) cnt -= 1.0;
    }
  }
  cout << setprecision(10) << cnt / all;
}