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;
int getint() {
  unsigned int c;
  int x = 0;
  while (((c = getchar()) - '0') >= 10) {
    if (c == '-') return -getint();
    if (!~c) exit(0);
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while (((c = getchar()) - '0') < 10);
  return x;
}
vector<long long> lucky;
const int M = 1e9 + 10;
void make(long long n) {
  if (n > M) return;
  lucky.push_back(n);
  make(n * 10 + 4);
  make(n * 10 + 7);
}
int main() {
  int i, j, k, tcc;
  int tc = 1 << 28;
  for (tcc = 0; tcc < tc; tcc++) {
    lucky.clear();
    make(0);
    long long pft = getint();
    long long prg = getint();
    long long vft = getint();
    long long vrg = getint();
    k = getint();
    sort((lucky).begin(), (lucky).end());
    lucky.push_back(4444444444LL);
    long long acount[1111] = {0};
    long long bcount[1111] = {0};
    for (i = 0; i < lucky.size() - 1; i++) {
      if (prg > lucky[i] and pft < lucky[i + 1]) {
        int ub = min(lucky[i + 1] - 1, (long long)prg);
        int lb = max(lucky[i] + 1, (long long)pft);
        acount[i] = ub - lb + 1;
      }
      if (vrg > lucky[i] and vft < lucky[i + 1]) {
        int ub = min(lucky[i + 1] - 1, (long long)vrg);
        int lb = max(lucky[i] + 1, (long long)vft);
        bcount[i] = ub - lb + 1;
      }
    }
    double suitable = 0;
    for (i = 0; i < 1111; i++)
      if (acount[i]) {
        for (j = 0; j < 1111; j++)
          if (bcount[j]) {
            if (abs(i - j) == k) {
              suitable += acount[i] * bcount[j];
            }
          }
      }
    for (i = 0; i < lucky.size(); i++)
      if (pft <= lucky[i] and lucky[i] <= prg) {
        int down = i - k;
        int up = i + k - 1;
        if (down >= 0) suitable += bcount[down];
        if (up < 1111) suitable += bcount[up];
      }
    for (i = 0; i < lucky.size(); i++)
      if (vft <= lucky[i] and lucky[i] <= vrg) {
        int down = i - k;
        int up = i + k - 1;
        if (down >= 0) suitable += acount[down];
        if (up < 1111) suitable += acount[up];
      }
    for (i = 0; i < lucky.size(); i++)
      if (pft <= lucky[i] and lucky[i] <= prg) {
        for (j = 0; j < lucky.size(); j++)
          if (vft <= lucky[j] and lucky[j] <= vrg) {
            int minc = min(i, j);
            int maxc = max(i, j);
            if (maxc - minc == k - 1) suitable++;
          }
      }
    double res = suitable / (prg - pft + 1) / (vrg - vft + 1);
    printf("%.10lf\n", res);
  }
  return 0;
}