How to get all the elements of an array that are divisible with 3 and then sum them thogether?

I dont know how to get the code to print out the elements that are divisable with 3 and the print out the sum of those elements , can someone help me do it , thanks for your time!

Code:

    #include <stdio.h>

int sum(int arr[]){
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 0;
    for (int y=0;y<n;y++){
        sum += arr[y];
        printf("%d",sum);
    }
}

int main() {
    int F[5] = {1,3,5,9,8};
    int s = 0;
     for (int i=0;i<5;i++){
       if (F[i]%3 == 0) {
           int diviz[] = {F[i]};
           printf("%d\n",diviz[0]);
           sum(diviz);
       }
    }
    return 0;
}

Expected Output:

3
9
12

Actual Output:

3
349
910

Idk how to solve this issue

>Solution :

int main() {
    int F[5] = {1,3,5,9,8};
    int s = 0;
     for (int i=0;i<5;i++){
       if (F[i]%3 == 0) {
           s = s + F[i];
           printf("%d",F[i]); 
       }
    }
    printf("%d",s); //print sum total 
    return 0;
}