Problem:
Score : 100 points

Problem Statement

You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

Constraints

0 \leq N_1, N_2, N_3, N_4 \leq 9

N_1, N_2, N_3 and N_4 are integers.

Input

Input is given from Standard Input in the following format:

N_1 N_2 N_3 N_4

Output

If N_1, N_2, N_3 and N_4 can be arranged into the sequence of digits "1974", print YES; if they cannot, print NO.

Sample Input 1

1 7 9 4

Sample Output 1

YES

We can get 1974 by swapping N_2 and N_3.

Sample Input 2

1 9 7 4

Sample Output 2

YES

We already have 1974 before doing anything.

Sample Input 3

1 2 9 1

Sample Output 3

NO

Sample Input 4

4 9 0 8

Sample Output 4

NO

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

 int main(void){
 char a[4];
 int i = 0;
 int bIs1 = 0;
 int bIs9 = 0;
 int bIs7 = 0;
 int bIs4 = 0;
 scanf("%s %s %s %s",&a[0],&a[1],&a[2],&a[3]);
   
 for(i =0; i < 4; i++) {
    if (a[i] == '1') {
       bIs1 = 1;
       break;
   }
}

 for(i =0; i < 4; i++) {
    if (a[i] == '9') {
       bIs9 = 1;
       break;
   }
}

  for(i =0; i < 4; i++) {
    if (a[i] == '7') {
       bIs7 = 1;
       break;
   }
}

  for(i =0; i < 4; i++) {
    if (a[i] == '4') {
       bIs4 = 1;
       break;
   }
}


 if(bIs1 * bIs9 * bIs7 * bIs4) {
    printf("YES\n");
} else {
   printf("NO\n");
}

return 0;
}