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> vec;
void dfs(int r, int now, int v) {
  if (now == r) {
    vec.push_back(v);
    return;
  }
  dfs(r, now + 1, v * 10 + 4);
  dfs(r, now + 1, v * 10 + 7);
}
long long pl, pr, vl, vr, K;
int size;
double cal() {
  double s = 0;
  for (int i = 0; i < size; i++) {
    long long lp = i == 0 ? 1 : vec[i - 1] + 1;
    lp = max(lp, pl);
    long long rp = vec[i];
    rp = min(rp, pr);
    if (rp < lp) continue;
    int rk = i + K - 1;
    if (rk >= size) continue;
    long long lv = vec[rk];
    lv = max(lv, vl);
    long long rv = rk == size - 1 ? 0x7fffffff : vec[rk + 1] - 1;
    rv = min(rv, vr);
    if (rv < lv) continue;
    s += (rp - lp + 1) * 1.0 * (rv - lv + 1) * 1.0;
  }
  return s;
}
bool check(int l, int r, int x) {
  if (l <= x && r >= x) return true;
  return false;
}
int main() {
  vec.clear();
  for (int i = 1; i <= 9; i++) {
    dfs(i, 0, 0);
  }
  sort(vec.begin(), vec.end());
  size = vec.size();
  while (cin >> pl >> pr >> vl >> vr >> K) {
    double sum = (pr - pl + 1) * 1.0 * (vr - vl + 1);
    double ans = 0;
    ans += cal();
    swap(pl, vl);
    swap(pr, vr);
    ans += cal();
    if (K == 1) {
      for (int i = 0; i < size; i++) {
        if (check(pl, pr, vec[i]) && check(vl, vr, vec[i])) ans--;
      }
    }
    printf("%.12f\n", ans / sum);
  }
  return 0;
}