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;
const int maxn = 1e6 + 100;
long long pl, pr, vl, vr, k;
vector<long long> g;
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 1; i <= 11; ++i) {
    int mx = (1 << i);
    for (int k = 0; k < mx; ++k) {
      long long now = 0;
      for (int j = 0; j < i; ++j) {
        now *= 10;
        if (k & (1 << j)) {
          now += 4;
        } else
          now += 7;
      }
      g.push_back(now);
    }
  }
  g.push_back(0);
  sort(g.begin(), g.end());
  long long ans = 0;
  for (int i = 1; i + k <= g.size(); ++i) {
    if (g[i] < pl) continue;
    if (g[i - 1] >= pr) break;
    long long now = max(g[i - 1] + 1, pl);
    long long have = min(pr, g[i]);
    long long last = have - now + 1;
    long long pos = i + k - 1;
    long long pos2 = pos + 1;
    if (vl < g[pos2] && vr >= g[pos]) {
      long long nmx = min(g[pos2] - 1, vr);
      long long nmn = max(g[pos], vl);
      long long have2 = nmx - nmn + 1;
      ans += 1ll * last * have2;
    }
  }
  for (int i = 1; i + k <= g.size(); ++i) {
    if (g[i] < vl) continue;
    if (g[i - 1] >= vr) break;
    long long now = max(g[i - 1] + 1, vl);
    long long have = min(vr, g[i]);
    long long last = have - now + 1;
    long long pos = i + k - 1;
    long long pos2 = pos + 1;
    if (pl < g[pos2] && pr >= g[pos]) {
      long long nmx = min(g[pos2] - 1, pr);
      long long nmn = max(g[pos], pl);
      long long have2 = nmx - nmn + 1;
      ans += 1ll * last * have2;
    }
  }
  if (k == 1) {
    for (int i = 1; i < g.size(); ++i) {
      if (vl <= g[i] && g[i] <= vr && pl <= g[i] && g[i] <= pr) {
        ans--;
      }
    }
  }
  long long sums = 1ll * (pr - pl + 1) * (vr - vl + 1);
  printf("%.12lf\n", 1.0 * ans / sums);
  return 0;
}