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 luck[2000];
int num;
void DFS(long long k) {
  if (k > 1000000000) return;
  luck[num] = k;
  num++;
  DFS(k * 10 + 4);
  DFS(k * 10 + 7);
}
int main() {
  long long pl, pr, vl, vr;
  int k;
  num = 0;
  DFS(0);
  sort(luck, luck + num);
  luck[num] = 1000000000;
  while (scanf("%I64d%I64d%I64d%I64d%d", &pl, &pr, &vl, &vr, &k) != EOF) {
    double sum = 0;
    for (int i = 1; i + k - 1 < num; i++) {
      double l, r;
      l = 0;
      r = 0;
      if (luck[i] >= pl && luck[i - 1] < pr) {
        if (pr > luck[i]) {
          if (pl <= luck[i - 1])
            l = luck[i] - luck[i - 1];
          else
            l = luck[i] - pl + 1;
        } else {
          if (pl <= luck[i - 1])
            l = pr - luck[i - 1];
          else
            l = pr - pl + 1;
        }
      }
      if (luck[i + k - 1] <= vr && luck[i + k] > vl) {
        if (vl < luck[i + k - 1]) {
          if (vr >= luck[i + k])
            r = luck[i + k] - luck[i + k - 1];
          else
            r = vr - luck[i + k - 1] + 1;
        } else {
          if (vr >= luck[i + k])
            r = luck[i + k] - vl;
          else
            r = vr - vl + 1;
        }
      }
      sum += l * r;
      l = 0;
      r = 0;
      if (luck[i] >= vl && luck[i - 1] < vr) {
        if (vr > luck[i]) {
          if (vl <= luck[i - 1])
            l = luck[i] - luck[i - 1];
          else
            l = luck[i] - vl + 1;
        } else {
          if (vl <= luck[i - 1])
            l = vr - luck[i - 1];
          else
            l = vr - vl + 1;
        }
      }
      if (luck[i + k - 1] <= pr && luck[i + k] > pl) {
        if (pl < luck[i + k - 1]) {
          if (pr >= luck[i + k])
            r = luck[i + k] - luck[i + k - 1];
          else
            r = pr - luck[i + k - 1] + 1;
        } else {
          if (pr >= luck[i + k])
            r = luck[i + k] - pl;
          else
            r = pr - pl + 1;
        }
      }
      sum += l * r;
      if (k == 1) {
        if (luck[i] >= pl && luck[i] <= pr && luck[i] >= vl && luck[i] <= vr)
          sum -= 1;
      }
    }
    double ans = (pr - pl + 1) * (vr - vl + 1);
    ans = sum / ans;
    printf("%.12f\n", ans);
  }
  return 0;
}