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;
long long modpow(long long a, long long b,
                 long long mod = (long long)(1e9 + 7)) {
  if (!b) return 1;
  a %= mod;
  return modpow(a * a % mod, b / 2, mod) * (b & 1 ? a : 1) % mod;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<long long> luck;
void genLucky() {
  luck.push_back(0);
  luck.push_back(4);
  while (luck.back() < 1e9) {
    long long x = luck.back();
    long long base = 1;
    long long nxt = 0;
    while (x) {
      if (x % 10 == 4) {
        nxt += (x + 3) * base;
        break;
      } else {
        nxt += 4 * base;
        x /= 10;
        if (!x) nxt += 4 * base * 10;
      }
      base *= 10;
    }
    luck.push_back(nxt);
  }
}
long long inter(long long a, long long b, long long l, long long r) {
  if (a > b || l > r)
    return 0;
  else if (a > r || b < l)
    return 0;
  else if (a > l)
    return inter(l, r, a, b);
  if (a <= l && r <= b) return r - l + 1;
  return b - l + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b, l, r;
  int k;
  genLucky();
  while (cin >> a >> b >> l >> r >> k) {
    long double num = 0;
    for (int i = 1; i <= luck.size() - k; i++) {
      int j = i + k - 1;
      num += inter(a, b, luck[i - 1] + 1, luck[i]) *
             inter(l, r, luck[j], luck[j + 1] - 1);
      num += inter(l, r, luck[i - 1] + 1, luck[i]) *
             inter(a, b, luck[j], luck[j + 1] - 1);
      if (k == 1)
        num -= inter(l, r, luck[i], luck[i]) * inter(a, b, luck[i], luck[i]);
    }
    cout << setprecision(12) << num / (b - a + 1) / (r - l + 1) << endl;
  }
  return 0;
}