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;
vector<int> vp;
int p_l, p_r, v_l, v_r;
inline void pre(void) {
  for (int i = 1; i <= 9; i++) {
    int max_mask = (1 << i);
    for (int mask = 0; mask < max_mask; mask++) {
      int curr = 0;
      int mul = 1;
      for (int j = 0; j < i; j++) {
        if (mask & (1 << j)) {
          curr += (mul * 7);
        } else {
          curr += (mul * 4);
        }
        mul *= 10;
      }
      vp.push_back(curr);
    }
  }
  sort(vp.begin(), vp.end());
}
int K;
inline long double power(long double a, int p) {
  long double ret = 1;
  while (p > 0) {
    if (p & 1) {
      ret *= a;
    }
    a *= a;
    p /= 2;
  }
  return ret;
}
int inrange(int a, int b, int l, int r) {
  int ret = min(b, r) - max(a, l) + 1;
  ret = max(ret, 0);
  return ret;
}
inline void ReadInput(void) {
  scanf("%d", &p_l);
  scanf("%d", &p_r);
  scanf("%d", &v_l);
  scanf("%d", &v_r);
  scanf("%d", &K);
}
inline void solve(void) {
  pre();
  double ans = 0;
  for (int i = 0; i < vp.size(); i++) {
    if (i + K > vp.size()) break;
    int prev = (i ? vp[i - 1] : 0);
    int next = (i + K < (int)vp.size() ? vp[i + K] : 1e9 + 1);
    int cnt1, cnt2;
    cnt1 = inrange(v_l, v_r, prev + 1, vp[i]);
    cnt2 = inrange(p_l, p_r, vp[i + K - 1], next - 1);
    double g, f;
    g = cnt1 / (double)(v_r - v_l + 1);
    f = cnt2 / (double)(p_r - p_l + 1);
    ans += (g * f);
    bool flag = false;
    if (cnt1 and cnt2) flag = true;
    cnt1 = inrange(p_l, p_r, prev + 1, vp[i]);
    cnt2 = inrange(v_l, v_r, vp[i + K - 1], next - 1);
    g = cnt1 / (double)(p_r - p_l + 1);
    f = cnt2 / (double)(v_r - v_l + 1);
    ans += (g * f);
    if (K == 1 and cnt1 and cnt2 and flag) {
      g /= cnt1;
      f /= cnt2;
      ans -= (g * f);
    }
  }
  printf("%.10lf\n", ans);
}
inline void Refresh(void) {}
int main() {
  ReadInput();
  solve();
  return 0;
}