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 INF = 0x7FFFFFFF;
const double eps = 1e-10;
const double pi = acos(-1.0);
vector<int> lucky;
int pl, pr, vl, vr, kth;
void init() { cin >> pl >> pr >> vl >> vr >> kth; }
void dfs(long long now) {
  if (now > max(pr, vr)) return;
  lucky.push_back(now);
  dfs(now * 10 + 4);
  dfs(now * 10 + 7);
}
void solve() {
  lucky.clear();
  dfs(4);
  dfs(7);
  sort(lucky.begin(), lucky.end());
  int le = 0, ri = kth - 1;
  if (ri + 1 > lucky.size() || lucky[ri] > max(pr, vr)) {
    cout << 0 << endl;
    return;
  }
  double ret = 0;
  double p = (double)(pr - pl + 1) * (vr - vl + 1);
  for (; ri < lucky.size(); ri++, le++) {
    if (lucky[le] >= pl && lucky[ri] <= vr && pl <= pr) {
      double x1, x2;
      x1 = min(pr - pl + 1, lucky[le] - pl + 1);
      if (ri == lucky.size() - 1) {
        x2 = min(vr - vl + 1, vr - lucky[ri] + 1);
      } else {
        if (vr == lucky[ri + 1])
          x2 = lucky[ri + 1] - lucky[ri];
        else
          x2 = min(vr, lucky[ri + 1] - 1) - max(lucky[ri], vl) + 1;
      }
      if (!(ri != lucky.size() - 1 && vl >= lucky[ri + 1])) ret += x1 * x2 / p;
    }
    if (lucky[le] >= vl && lucky[ri] <= pr && vl <= vr) {
      double x1, x2;
      x1 = min(vr - vl + 1, lucky[le] - vl + 1);
      if (ri == lucky.size() - 1) {
        x2 = min(pr - pl + 1, pr - lucky[ri] + 1);
      } else {
        if (pr == lucky[ri + 1])
          x2 = lucky[ri + 1] - lucky[ri];
        else
          x2 = min(pr, lucky[ri + 1] - 1) - max(pl, lucky[ri]) + 1;
      }
      if (pl <= lucky[le] && lucky[le] <= pr && kth == 1)
        ret += (x1 * x2 - 1) / p;
      else if (!(ri != lucky.size() - 1 && pl >= lucky[ri + 1]))
        ret += x1 * x2 / p;
    }
    if (lucky[le] >= pl) pl = lucky[le] + 1;
    if (lucky[le] >= vl) vl = lucky[le] + 1;
  }
  printf("%.10f\n", ret);
}
int main() {
  init();
  solve();
  return 0;
}