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 int N = 55;
const long long M = 1000000000;
const int K = 670000;
const int LIT = 2500;
const int INF = 1 << 30;
const int base = 137;
const double eps = 1e-10;
const int dir[5][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
const int dir2[10][2] = {{-1, 0}, {1, 0},  {0, 1},  {0, -1},
                         {1, 0},  {-1, 0}, {0, -1}, {0, 1}};
const int ABS(int a) { return a > 0 ? a : -a; }
vector<long long> lk;
long long p1, p2, v1, v2, k;
void dfs(long long u) {
  if (u > M) return;
  lk.push_back(u);
  dfs(u * 10 + 4);
  dfs(u * 10 + 7);
}
void init() {}
pair<long long, long long> getins(pair<long long, long long> a,
                                  pair<long long, long long> b) {
  if (a.first > b.first) swap(a, b);
  long long lt, rt;
  if (a.second < b.first)
    lt = -1, rt = -1;
  else
    lt = max(a.first, b.first), rt = min(a.second, b.second);
  return make_pair(lt, rt);
}
long long cal() {
  long long ret = 0;
  int s = lk.size();
  for (int i = 1; i < s; i++) {
    if (i + k - 1 >= s) break;
    long long bnd = (i + k < s) ? lk[i + k] - 1 : M;
    pair<long long, long long> lt =
        getins(make_pair(lk[i - 1] + 1, lk[i]), make_pair(p1, p2));
    pair<long long, long long> rt =
        getins(make_pair(lk[i + k - 1], bnd), make_pair(v1, v2));
    if (lt.first == -1 || rt.first == -1) continue;
    long long tmp = (lt.second - lt.first + 1) * (rt.second - rt.first + 1);
    ret += tmp;
  }
  return ret;
}
long long spc() {
  pair<long long, long long> gap = getins(make_pair(p1, p2), make_pair(v1, v2));
  if (gap.first == -1) return 0;
  int s = lk.size();
  long long ret = 0;
  for (int i = 1; i < s; i++) {
    if (gap.first <= lk[i] && gap.second >= lk[i]) ret++;
  }
  return ret;
}
void solve() {
  long long base = (p2 - p1 + 1) * (v2 - v1 + 1);
  long long cnt = 0;
  cnt += cal();
  swap(p1, v1);
  swap(p2, v2);
  cnt += cal();
  if (k == 1) cnt -= spc();
  double res = (double)cnt / (double)base;
  printf("%.10lf\n", res);
}
int main() {
  dfs(0);
  sort(lk.begin(), lk.end());
  while (cin >> p1 >> p2 >> v1 >> v2 >> k) {
    init();
    solve();
  }
}