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;
long long pr, pl, vl, vr, k;
set<long long> my;
vector<long long> R;
void made(long long x) {
  if (x >= 1e10) {
    return;
  }
  long long x1 = x * 10 + 4;
  long long x2 = x * 10 + 7;
  made(x1);
  made(x2);
  my.insert(x1);
  my.insert(x2);
}
double memo[2005][2005][2];
int vis[2005][2005][2];
double solve(int i, int j, int ok) {
  if (i > j) {
    return 0;
  }
  double &ret = memo[i][j][ok];
  if (vis[i][j][ok]) return ret;
  vis[i][j][ok] = 1;
  ret = 0;
  if (j - i + 1 == k) {
    if (R[i] >= pl) {
      double x1 = R[i] - R[i - 1];
      x1 = min(x1, 1.0 * min(R[i], pr) - max(R[i - 1], pl - 1));
      x1 = max(x1, 0.0);
      if (R[j] <= vr) {
        double x2 = R[j + 1] - R[j];
        x2 = min(x2, 1.0 * min(R[j + 1], vr + 1) - max(R[j], vl));
        x2 = max(x2, 0.0);
        ret += x1 * x2;
      }
    }
    if (R[i] >= vl) {
      double x1 = R[i] - R[i - 1];
      x1 = min(x1, 1.0 * min(R[i], vr) - max(R[i - 1], vl - 1));
      x1 = max(x1, 0.0);
      if (R[j] <= pr) {
        double x2 = R[j + 1] - R[j];
        x2 = min(x2, 1.0 * min(R[j + 1], pr + 1) - max(R[j], pl));
        x2 = max(x2, 0.0);
        if (k == 1 && x1 + x2 > 0) {
          ret--;
        }
        ret += x1 * x2;
      }
    }
  }
  if (!ok) ret += solve(i + 1, j, 0);
  ret += solve(i, j - 1, 1);
  return ret;
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  double ans = 1.0 / (1.0 * (pr - pl + 1) * (vr - vl + 1));
  made(0);
  my.insert(0);
  for (auto x : my) {
    if (R.size() > 1500) break;
    R.push_back(x);
  }
  double ans2 = solve(1, R.size() - 2, 0);
  printf("%.9f", (ans * ans2));
}