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.
The first line contains two integers *n* and *k* (1 ≤ *n*, *k* ≤ 10<sup class="upper-index">9</sup>) — the number of elements in the permutation and the lexicographical number of the permutation.
7 4
1
4 7
1
7 1
2
7 5040
1
10 1023
0
7 7477
-1
10 10000
1
3 7
-1
27 1
2
40 8544
2
47 1
4
47 8547744
3
50 1000000000
4
64 87
4
98 854555
6
100 1
6
9985 5888454
30
1 1
0
1 2
-1
2 1000000000
-1
10 1000000000
-1
20 1000000000
2
777777 1
126
777777 2
125
777474 10000
120
1000000000 1
1022
777777777 5
1021
777777777 1
1022
777477774 1
989
444747744 1000000000
554
475 88555458
8
12 855448
1
20 1000000000
2
47 99998544
2
49 1000000000
2
854459 95554455
126
77777779 1000000000
508
77 47
5
6999 85488877
22
7479 58884598
24
1000000000 1000000000
1022
7 1000
0
7 124
1
7 2048
1
7 3001
1
7 127
1
7 980
1
7 5000
0
7 4095
2
7 3856
0
7 5032
0
7 4999
0
2 3
-1
2 4
-1
7 985
1
4 25
-1
6 121
1
11 39916801
-1
29 1000000000
2
10 4589
1
10 100000
1
10 98564
1
10 1
2
10 100000009
-1
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
int y,m,d;
int shift;
int m1[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int m2[15]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int main()
{
	scanf("%d.%d.%d\n%d",&d,&m,&y,&shift);
	int k=365*3+366;
	int day=d+(y/4)*k+(y%4)*365;
	for(int i=1;i<m;i++)
		day+=y%4?m1[i]:m2[i];
	day+=shift;
	if(shift<0)
		day++;
	int ny=0;
	while(1)
	{
		if(ny%4==0&&day>366)
		{
			day-=366;
			ny++;
		}
		else if(ny%4&&day>365)
		{
			day-=365;
			ny++;
		}
		else
			break;
	}
	int nd=day;
	int nm=1;
	if(ny%4==0)
	{
		for(int i=1;i<=12;i++)
		{
			if(nd<=m2[i])
			{
				nm=i;
				break;
			}
			else
				nd-=m2[i];
		}
	}
	else
	{
		for(int i=1;i<=12;i++)
		{
			if(nd<=m1[i])
			{
				nm=i;
				break;
			}
			else
				nd-=m1[i];
		}
	}
	printf("%02d.%02d.%d\n",nd,nm,ny);
	return 0;
}
