Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

One day Petya dreamt of a lexicographically k-th permutation of integers from 1 to n. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 109) — the number of elements in the permutation and the lexicographical number of the permutation.

Output

If the k-th permutation of numbers from 1 to n does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes i, that i and ai are both lucky numbers.

Examples

Input

7 4


Output

1


Input

4 7


Output

1

Note

A permutation is an ordered set of n elements, where each integer from 1 to n occurs exactly once. The element of permutation in position with index i is denoted as ai (1 ≤ i ≤ n). Permutation a is lexicographically smaller that permutation b if there is such a i (1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj. Let's make a list of all possible permutations of n elements and sort it in the order of lexicographical increasing. Then the lexicographically k-th permutation is the k-th element of this list of permutations.

In the first sample the permutation looks like that:

1 2 3 4 6 7 5

The only suitable position is 4.

In the second sample the permutation looks like that:

2 1 3 4

The only suitable position is 4.
import java.util.*;

public class EPerm {
	static int[] factorial = {-1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600};
	static long[] luckyNums = new long[1024];
	static HashSet<Long> theNums = new HashSet<Long>(5000);
	static int[] luckyList = {4, 7};
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int fact = getFact(k);
		int[] lastPart = new int[fact];
		for(int i = 0; i < lastPart.length; ++i) {
			lastPart[i] = n - fact + i + 1;
		}
		makeLucky();
		--k;
		if(fact > n)
			System.out.println("-1");
		else {
			int index = 0;
			while(k > 0 && fact > 1) {
				--fact;
				int multiple = k/(factorial[fact]);
				if(multiple == fact + 1)
					--multiple;
				k -= multiple*factorial[fact];
				int temp = lastPart[multiple+index];
				lastPart[multiple+index] = lastPart[index];
				lastPart[index] = temp;
				++index;
				Arrays.sort(lastPart, index, lastPart.length);
			}
			
			int total = 0;
			int save = n - lastPart.length;
			total += getCount(save); //inclusive
			//System.out.println("Total is now " + total);
			//then iterate through the array lastPart to check if anything is a lucky num match
			for(int i = 0; i < lastPart.length; ++i) {
				//check if both i + (n - lastPart.length) and lastPart[i + ..] are luckyNums
				//System.out.printf("Checking %d and %d\n", i + save+1, lastPart[i]);
				if( theNums.contains((long)(i + save+1)) && theNums.contains((long)lastPart[i]) )
					++total;
			}
			System.out.println(total);
		}
		
	}
	
	public static int getCount(int n) {
		int ret = Arrays.binarySearch(luckyNums, n);
		if(ret < 0)
			ret = -ret - 2;
		return ret;
	}
	
	public static int getFact(int k) {
		int ret = Arrays.binarySearch(factorial, k);
		if(ret < 0)
			ret = -ret-1;
		return ret;
	}
	
	public static void makeLucky() {
		int prevStart = 0, prevEnd = 0;
		int curIndex = 1;
		for(int len = 1; len <= 9; ++len) {
			for(int i = prevStart; i <= prevEnd; ++i) {
				for(int poss: luckyList) {
					luckyNums[curIndex] = luckyNums[i]*10 + poss;
					theNums.add(luckyNums[curIndex]);
					++curIndex;
				}
			}
			prevStart = prevEnd + 1;
			prevEnd = curIndex - 1;
		}
		luckyNums[1023] = 4444444444L;
		theNums.add(luckyNums[1023]);
	}
}
