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 double PI = atan(1.0) * 4;
const int64_t INF = 100000000000000003;
const int32_t LOG = 21;
vector<long long int> vals[12];
vector<long long int> v;
void lucky() {
  long long int cur = 0;
  long long int i = 1;
  vals[0].push_back(0);
  while (i < 10) {
    for (long long int j = 0; j < vals[i - 1].size(); j++) {
      long long int no1 = vals[i - 1][j] * 10 + 4,
                    no2 = vals[i - 1][j] * 10 + 7;
      v.push_back(no1);
      v.push_back(no2);
      vals[i].push_back(no1);
      vals[i].push_back(no2);
    }
    i++;
  }
  v.push_back(0);
  v.push_back(1e9 + 7);
  sort(v.begin(), v.end());
}
long long int intersect(long long int l1, long long int r1, long long int l2,
                        long long int r2) {
  if (l1 > l2) {
    swap(l1, l2);
    swap(r1, r2);
  }
  if (l1 <= l2 && r2 <= r1) return r2 - l2 + 1;
  return max(r1 - l2 + 1, 0LL);
}
double solve(long long int l1, long long int r1, long long int dist1,
             long long int l2, long long int r2, long long int dist2,
             long long int k) {
  if (l1 > r1 || l2 > r2) return 0;
  long long int n = v.size() - 1;
  double ans = 0.0000;
  for (long long int i = 1; i < n; i++) {
    long long int j = i + k - 1;
    if (j >= n) break;
    long long int range1_l = v[i - 1] + 1;
    long long int range1_r = v[i];
    long long int range2_l = v[j];
    long long int range2_r = v[j + 1] - 1;
    double val1 = intersect(range1_l, range1_r, l1, r1);
    double val2 = intersect(range2_l, range2_r, l2, r2);
    val1 = val1 / dist1;
    val2 = val2 / dist2;
    double toAdd = val1 * val2;
    ans += toAdd;
  }
  return ans;
}
double count1(long long int l, long long int r) {
  long long int ans = 0;
  for (long long int i = 1; i < v.size() - 1; i++) {
    if (v[i] >= l && v[i] <= r) ans++;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  lucky();
  long long int l1, r1, l2, r2, k;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  if (l1 > l2 || (l1 == l2 && r1 > r2)) {
    swap(l1, l2);
    swap(r1, r2);
  }
  long long int dist1 = r1 - l1 + 1, dist2 = r2 - l2 + 1;
  double ans = 0.000000;
  if (l2 <= r1 && r1 < r2) {
    ans += solve(l1, l2 - 1, dist1, r1 + 1, r2, dist2, k);
    ans += solve(l1, l2 - 1, dist1, l2, r1, dist2, k);
    ans += solve(l2, r1, dist1, r1 + 1, r2, dist2, k);
    ans += 2.0000 * solve(l2, r1, dist1, l2, r1, dist2, k);
    if (k == 1) ans -= count1(l2, r1) / (dist2 * dist1);
  } else if (l2 <= r1 && r2 <= r1) {
    ans += solve(l1, l2 - 1, dist1, l2, r2, dist2, k);
    ans += solve(l2, r2, dist2, r2 + 1, r1, dist1, k);
    ans += 2 * solve(l2, r2, dist1, l2, r2, dist2, k);
    if (k == 1) ans -= count1(l2, r2) / (dist2 * dist1);
  } else {
    ans += solve(l1, r1, dist1, l2, r2, dist2, k);
  }
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}