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;
int p1, p2, v1, v2, k;
int minVal, maxVal;
double res, total;
void gen(long long val) {
  if (val > maxVal) return;
  if (val >= minVal && val <= maxVal) lucky.push_back(val);
  gen(val * 10 + 4);
  gen(val * 10 + 7);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  total = (double)(p2 - p1 + 1) * (v2 - v1 + 1);
  minVal = min(p1, v1);
  maxVal = max(p2, v2);
  gen(0);
  sort(lucky.begin(), lucky.end());
  for (int i = 0; i + k - 1 < (int)lucky.size(); i++) {
    int j = i + k - 1;
    int minL, maxL = lucky[i];
    if (i == 0)
      minL = minVal;
    else
      minL = lucky[i - 1] + 1;
    int minR = lucky[j], maxR;
    if (j == (int)lucky.size() - 1)
      maxR = maxVal;
    else
      maxR = lucky[j + 1] - 1;
    double a1, b1, c1, d1;
    a1 = max(minL, p1), b1 = min(maxL, p2);
    c1 = max(minR, v1), d1 = min(maxR, v2);
    if (a1 <= b1 && c1 <= d1) res += (b1 - a1 + 1) * (d1 - c1 + 1) / total;
    double a2, b2, c2, d2;
    c2 = max(minL, v1), d2 = min(maxL, v2);
    a2 = max(minR, p1), b2 = min(maxR, p2);
    if (a2 <= b2 && c2 <= d2) res += (b2 - a2 + 1) * (d2 - c2 + 1) / total;
    double a3 = max(a1, a2), b3 = min(b1, b2);
    double c3 = max(c1, c2), d3 = min(d1, d2);
    if (a3 <= b3 && c3 <= d3) res -= (b3 - a3 + 1) * (d3 - c3 + 1) / total;
  }
  cout << fixed << setprecision(9) << res << endl;
  return 0;
}