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)

#include <stdio.h>

int main() {
    int i, j, tmp, N1, N2, N3, N4;
    int array[4];
    
    scanf("%d", &N1);
    scanf("%d", &N2);
    scanf("%d", &N3);
    scanf("%d", &N4);
    
    array[0] = N1, array[1] = N2, array[2] = N3, array[3] = N4;
    
    /* 数値を降順にソート */
    for (i = 0; i < 4; ++i) {
        for (j = i + 1; j < 4; ++j) {
            if (array[i] < array[j]) {
                tmp =  array[i];
                array[i] = array[j];
                array[j] = tmp;
            }
        }
    }
    
    /* 降順にソートして，9741であるか確認 */
    if (array[0] == 9 && array[1] == 7 && array[2] == 4 && array[3] == 1) {
        printf("YES");
    }else{
        printf("NO");
    }

    
    return 0;
}