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>
const int N = 2e3 + 9;
const int inf = 1e9 + 7;
using namespace std;
using ll = long long;
int num[N], cnt, pl, pr, vl, vr, k;
ll ans = 0;
void dfs(ll s) {
  if (s > 1e9) return;
  if (s) num[cnt++] = s;
  dfs(s * 10 + 4);
  dfs(s * 10 + 7);
}
ll slove(int L, int l, int r, int R) {
  ll res = 0;
  res += 1LL * max(min(pr, l) - max(pl, L + 1) + 1, 0) *
         max(min(R - 1, vr) - max(vl, r) + 1, 0);
  res += 1LL * max(min(vr, l) - max(vl, L + 1) + 1, 0) *
         max(min(R - 1, pr) - max(pl, r) + 1, 0);
  if (l == r && max(pl, vl) <= l && l <= min(pr, vr)) res--;
  return res;
}
int main() {
  dfs(0);
  sort(num, num + cnt);
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  for (int i = 0; i <= cnt - k && max(vr, pr) >= num[i + k - 1]; i++) {
    ans += slove(i > 0 ? num[i - 1] : 0, num[i], num[i + k - 1],
                 (i + k) < cnt ? num[i + k] : inf);
  }
  return printf("%.12f\n", 1.0 * ans / (pr - pl + 1) / (vr - vl + 1)), 0;
}