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<long long> v;
long long vr, vl, pl, pr, k;
void dfs(long long x) {
  if (x >= 10000000000000LL) return;
  v.push_back(x);
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
long long getans(long long sx, long long sy, long long tx, long long ty) {
  if (tx <= sx && sy <= ty) return sy - sx + 1;
  if (sx <= tx && ty <= sy) return ty - tx + 1;
  if (sx > ty) return 0LL;
  if (tx > sy) return 0LL;
  return max(0LL, min(ty, sy) - max(sx, tx) + 1);
}
int main() {
  v.push_back(0);
  dfs(4);
  dfs(7);
  sort(v.begin(), v.end());
  cin >> pl >> pr >> vl >> vr >> k;
  long long res = 0;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] > pr && v[i] > vr) break;
    if (i + k - 1 >= v.size()) break;
    long long nxt = v[i + k - 1];
    long long tl = v[i - 1] + 1;
    long long tr = v[i + k] - 1;
    long long suml = getans(tl, v[i], pl, pr);
    long long sumr = getans(v[i + k - 1], tr, vl, vr);
    res += suml * sumr;
    suml = getans(tl, v[i], vl, vr);
    sumr = getans(v[i + k - 1], tr, pl, pr);
    res += suml * sumr;
  }
  if (k == 1) {
    for (int i = 0; i < v.size(); i++)
      if (pl <= v[i] && v[i] <= pr && vl <= v[i] && v[i] <= vr) res--;
  }
  printf("%.12f\n", (double(res) / (pr - pl + 1)) / (vr - vl + 1));
  return 0;
}