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;
int pl, pr, vl, vr, k;
int L = 1, M = 1000000000 + 1;
vector<int> v;
int unilen(int a, int b, int c, int d) {
  if (b < c or d < a) return 0;
  return min(b, d) - max(a, c) + 1;
}
int main() {
  v.push_back(0);
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 1; i <= 9; i++) {
    int d = (1 << i);
    for (int j = 0; j < (int)(d); ++j) {
      int t = 0;
      for (int k = 0; k < (int)(i); ++k) {
        t *= 10;
        if (j & (1 << (i - 1 - k)))
          t += 7;
        else
          t += 4;
      }
      v.push_back(t);
    }
  }
  long long s = 0;
  v.push_back(M);
  for (int i = 0; i < (int)(v.size() - 1); ++i) assert(v[i] < v[i + 1]);
  for (int i = 0; i < (int)(v.size()); ++i) {
    if (i == 0) continue;
    int j = i + k;
    if (j >= (int)v.size()) break;
    s += (long long)unilen(v[i - 1] + 1, v[i], pl, pr) *
         unilen(v[j - 1], v[j] - 1, vl, vr);
    s += (long long)unilen(v[i - 1] + 1, v[i], vl, vr) *
         unilen(v[j - 1], v[j] - 1, pl, pr);
    if (k == 1 and pl <= v[i] and v[i] <= pr and vl <= v[i] and v[i] <= vr) s--;
  }
  long long P = (long long)(pr - pl + 1) * (vr - vl + 1);
  printf("%.12f", (double)s / P);
  return 0;
}