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> lucky;
map<int, int> Mp;
bool islucky(int x) {
  int d;
  while (x) {
    d = x % 10;
    if (d != 4 && d != 7) return false;
    x /= 10;
  }
  return true;
}
void gen(int n) {
  if (n < 0 || n * 10 > 1000000000 || !islucky(n)) return;
  if (n) lucky.push_back(n);
  gen(n * 10 + 4);
  gen(n * 10 + 7);
}
void pre() {
  lucky.clear(), Mp.clear();
  gen(0);
  sort(lucky.begin(), lucky.end());
  int i;
  for (i = 0; i < lucky.size(); ++i) {
    Mp[lucky[i]] = i;
  }
}
long long count(int l1, int r1, int l2, int r2, int K) {
  int i, j;
  long long ret = 0;
  int st1, nd1, st2, nd2;
  for (i = 0; i < lucky.size(); ++i) {
    if (lucky[i] < l1) continue;
    if (i && r1 < lucky[i - 1]) break;
    st1 = l1;
    if (i) st1 = ((l1) > (lucky[i - 1] + 1) ? (l1) : (lucky[i - 1] + 1));
    nd1 = ((lucky[i]) < (r1) ? (lucky[i]) : (r1));
    j = i + K - 1;
    if (j >= lucky.size()) break;
    st2 = lucky[j];
    if (j + 1 < lucky.size())
      nd2 = lucky[j + 1] - 1;
    else
      nd2 = r2;
    st2 = ((st2) > (l2) ? (st2) : (l2));
    nd2 = ((nd2) < (r2) ? (nd2) : (r2));
    if (st2 > nd2) continue;
    ret += (long long)(nd1 - st1 + 1) * (nd2 - st2 + 1);
  }
  return ret;
}
int main() {
  pre();
  int l1, r1, l2, r2, K, i;
  long long ans;
  while (scanf("%d %d %d %d %d", &l1, &r1, &l2, &r2, &K) != EOF) {
    ans = count(l1, r1, l2, r2, K) + count(l2, r2, l1, r1, K);
    if (K == 1) {
      for (i = 0; i < lucky.size(); ++i) {
        if (l1 <= lucky[i] && lucky[i] <= r1 && l2 <= lucky[i] &&
            lucky[i] <= r2)
          --ans;
      }
    }
    printf("%.11lf\n", (double)ans / ((double)(r1 - l1 + 1) * (r2 - l2 + 1)));
  }
  return 0;
}