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 int MAX_K = 9;
const int KBITS = 1 << MAX_K;
const int MAX_N = KBITS * 2 + 2;
const int INF = 1 << 30;
int lns[MAX_N];
int main() {
  int n = 0;
  lns[n++] = 0;
  for (int k = 1; k <= MAX_K; k++) {
    int kbits = 1 << k;
    for (int bits = 0; bits < kbits; bits++) {
      int sum = 0;
      for (int i = 0, bi = 1, t = 1; i < k; i++, bi <<= 1, t *= 10)
        sum += ((bits & bi) ? 7 : 4) * t;
      lns[n++] = sum;
    }
  }
  lns[n++] = INF;
  int p0, p1, v0, v1, k;
  scanf("%d%d%d%d%d", &p0, &p1, &v0, &v1, &k);
  p1++, v1++;
  long long all = (long long)(p1 - p0) * (v1 - v0);
  long long sum = 0;
  for (int i = 0, j = k; j < n; i++, j++) {
    int li0 = lns[i] + 1, li1 = lns[i + 1] + 1;
    int &lj0 = lns[j], &lj1 = lns[j + 1];
    int p00 = 0, p01 = 0, v00 = 0, v01 = 0;
    if (li0 < p1 && p0 < li1 && lj0 < v1 && v0 < lj1) {
      p00 = max(li0, p0), p01 = min(li1, p1);
      v00 = max(lj0, v0), v01 = min(lj1, v1);
      sum += (long long)(p01 - p00) * (v01 - v00);
    }
    int p10 = 0, p11 = 0, v10 = 0, v11 = 0;
    if (lj0 < p1 && p0 < lj1 && li0 < v1 && v0 < li1) {
      p10 = max(lj0, p0), p11 = min(lj1, p1);
      v10 = max(li0, v0), v11 = min(li1, v1);
      sum += (long long)(p11 - p10) * (v11 - v10);
    }
    if (p00 < p11 && p10 < p01 && v00 < v11 && v10 < v01) {
      int p20 = max(p00, p10), p21 = min(p01, p11);
      int v20 = max(v00, v10), v21 = min(v01, v11);
      sum -= (long long)(p21 - p20) * (v21 - v20);
    }
  }
  printf("%.12Lf\n", (long double)sum / all);
  return 0;
}