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 double eps = 1e-12;
int a[3100], n, m;
double inv[3100];
int lef[3100], rig[3100], p[3100], v[3100];
int pl, pr, vl, vr, k;
int num;
const int inf = 0x3fffffff;
void build(const int &x) {
  if (x)
    a[n++] = num;
  else
    n = num = 0;
  if (x == 9) return;
  num = num * 10 + 4;
  build(x + 1);
  num = num / 10;
  num = num * 10 + 7;
  build(x + 1);
  num = num / 10;
}
void init() {
  m = 0;
  for (int i = 0; i <= n; i++) {
    inv[m] = ((double)i) + 0.51;
    if (i != 0)
      lef[m] = a[i - 1] + 1;
    else
      lef[m] = 0;
    rig[m] = a[i] - 1;
    m++;
    if (i == n) break;
    inv[m] = ((double)i) + 1.00;
    lef[m] = rig[m] = a[i];
    m++;
  }
}
void work() {
  memset(p, 0, sizeof(p));
  memset(v, 0, sizeof(v));
  for (int i = 0; i < m; i++) {
    if (!(lef[i] > pr || rig[i] < pl)) {
      int l = max(lef[i], pl), r = min(rig[i], pr);
      p[i] = r - l + 1;
    }
    if (!(lef[i] > vr || rig[i] < vl)) {
      int l = max(lef[i], vl), r = min(rig[i], vr);
      v[i] = r - l + 1;
    }
  }
  double ans = 0.0;
  for (int i = 0; i < m; i++) {
    if (!p[i]) continue;
    for (int j = 0; j < m; j++) {
      if (!v[j]) continue;
      if (((((int)(inv[j])) - round(inv[i]) + 1) == k) ||
          ((((int)(inv[i])) - round(inv[j]) + 1) == k))
        ans += ((double)(p[i])) * ((double)(v[j]));
    }
  }
  ans /= ((double)(pr - pl + 1)) * ((double)(vr - vl + 1));
  printf("%.10f\n", ans);
}
int main() {
  build(0);
  sort(a, a + n);
  a[n] = inf;
  init();
  while (scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k) != EOF) work();
  return 0;
}