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;
void read_file(bool outToFile = true) {}
int pl, pr, vl, vr;
int K;
vector<int> lucky;
const int MAXV = 1000 * 1000 * 1000 + 1;
int lsz;
void gen(long long cur = 0) {
  if (cur >= MAXV) return;
  lucky.push_back(cur);
  gen(cur * 10 + 4);
  gen(cur * 10 + 7);
}
int len(int l, int r) { return max(0, r - l + 1); }
int calc(int l1, int r1, int l2, int r2) {
  int l = max(l1, l2);
  int r = min(r1, r2);
  return len(l, r);
}
bool in(int l, int pl, int pr) { return pl <= l && l <= pr; }
bool is_in(int l) { return in(l, vl, vr) && in(l, pl, pr); }
int main() {
  read_file();
  gen();
  lucky.push_back(MAXV);
  sort(((lucky).begin()), ((lucky).end()));
  lsz = lucky.size();
  while (scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &K) != -1) {
    int L = 1, R = K;
    long long ans = 0;
    while (R != lsz - 1) {
      int cnt1 = calc(lucky[L - 1] + 1, lucky[L], vl, vr);
      int cnt2 = calc(lucky[R], lucky[R + 1] - 1, pl, pr);
      long long nw = 1LL * cnt1 * cnt2;
      cnt1 = calc(lucky[L - 1] + 1, lucky[L], pl, pr);
      cnt2 = calc(lucky[R], lucky[R + 1] - 1, vl, vr);
      nw += 1LL * cnt1 * cnt2;
      if (L == R) {
        int l = lucky[L];
        if (is_in(l)) {
          assert(nw >= 2);
          nw--;
        }
      }
      ans += nw;
      L++, R++;
    }
    long long den = len(vl, vr) * 1LL * len(pl, pr);
    long double dans = ans * 1. / den;
    cout << setprecision(12) << fixed;
    cout << dans << '\n';
  }
}