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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int vl, vr, pl, pr, k;
vector<int> a;
void bfs() {
  queue<int> que;
  que.push(0);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (u > 1e8) continue;
    int tmp = u * 10 + 4;
    if (tmp > 1e9 || tmp < 0)
      ;
    else {
      que.push(tmp);
      a.push_back(tmp);
    }
    tmp = u * 10 + 7;
    if (tmp > 1e9 || tmp < 0)
      ;
    else {
      que.push(tmp);
      a.push_back(tmp);
    }
  }
}
int main() {
  a.push_back(0);
  bfs();
  a.push_back(1000000005);
  while (scanf("%d%d%d%d%d", &vl, &vr, &pl, &pr, &k) != EOF) {
    unsigned long long tot = (unsigned long long)(vr - vl + 1) * (pr - pl + 1);
    unsigned long long tmp = 0, cnt = 0;
    if (k == 1) {
      for (int i = 1; i < a.size() - 1; i++)
        if (a[i] >= vl && a[i] <= vr && a[i] >= pl && a[i] <= pr) cnt++;
    }
    for (int i = 1; i < a.size() - k + 1 - 1; i++) {
      if (a[i] < vl) continue;
      if (a[i - 1] >= vr) continue;
      if (a[i + k - 1] > pr) continue;
      if (a[i + k] <= pl) continue;
      tmp += (unsigned long long)(min(vr, a[i]) - max(a[i - 1] + 1, vl) + 1) *
             (min(pr, a[i + k] - 1) - max(pl, a[i + k - 1]) + 1);
    }
    for (int i = 1; i < a.size() - k + 1 - 1; i++) {
      if (a[i] < pl) continue;
      if (a[i - 1] >= pr) continue;
      if (a[i + k - 1] > vr) continue;
      if (a[i + k] <= vl) continue;
      tmp += (unsigned long long)(min(pr, a[i]) - max(a[i - 1] + 1, pl) + 1) *
             (min(vr, a[i + k] - 1) - max(vl, a[i + k - 1]) + 1);
    }
    printf("%.10f\n", (tmp - cnt) * 1.0 / tot);
  }
  return 0;
}