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 long long INF = 1e11 + 7;
int pl, pr, vl, vr, k;
vector<long long> LN;
void init() {
  queue<long long> que;
  que.push(0);
  while (!que.empty()) {
    long long v = que.front();
    que.pop();
    if (v > 0) LN.push_back(v);
    v *= 10;
    if (v + 4 < INF) que.push(v + 4);
    if (v + 7 < INF) que.push(v + 7);
  }
  sort((LN).begin(), (LN).end());
}
double get(long long l1, long long r1, long long l2, long long r2) {
  if (r1 < l2 || r2 < l1) return 0;
  return min(r1, r2) - max(l1, l2) + 1;
}
double solve() {
  double ret = 0;
  for (int i = 0; i + k - 1 < ((int)(LN).size()); ++i) {
    long long l, r;
    double p1, p2;
    l = i ? LN[i - 1] + 1 : 1, r = LN[i];
    p1 = get(l, r, pl, pr) / (pr - pl + 1);
    l = LN[i + k - 1], r = i + 1 < ((int)(LN).size()) ? LN[i + k] - 1 : INF;
    p2 = get(l, r, vl, vr) / (vr - vl + 1);
    ret += p1 * p2;
    l = i ? LN[i - 1] + 1 : 1, r = LN[i];
    p2 = get(l, r, vl, vr) / (vr - vl + 1);
    l = LN[i + k - 1], r = i + 1 < ((int)(LN).size()) ? LN[i + k] - 1 : INF;
    p1 = get(l, r, pl, pr) / (pr - pl + 1);
    ret += max(0., p1 * p2 - 1.0 * (k == 1) / (pr - pl + 1) / (vr - vl + 1));
  }
  return ret;
}
int main() {
  init();
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  printf("%.15f", solve());
  return 0;
}