scanf function with MPI

Question

I'm learning MPI and was fiddling with some of the besic functions, like MPI_Bcast. Usually one has the message in the main rank (i.e. 0) and broadcasts it to the others, so something like (this is a basic example from my notes)

#include "mpi.h"
#include <iostream>

int main(int argc, char* argv[]){

    // common buffer 'n' for every process
    int n, myid, numprocs, i;

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
    MPI_Comm_rank(MPI_COMM_WORLD, &myid);

    // at first only process 0 has the value of n, e.g. n=5
    if(myid==0){
        printf("Enter n: ");
        scanf("%d", &n);
    }

    // broadcasted to every other process, now each of
    // has a separate n that can be independently modified
    MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);

    if(myid==0){
        ++n; // 5->6 for proc 0
    }

    if(myid==1){
        n+=2; // 5->7 for proc 1
    }

    printf("n from process %d is %d\n", myid, n);
    /*
    upon execution (-np 4): 
    
    Enter n: 5
    n from process 0 is 6
    n from process 1 is 7
    n from process 2 is 5
    n from process 3 is 5
    */

    return 0;
}

Out of curiosity i tried to put the scanf out of the if clause, so to give every process an initial value for n:

#include "mpi.h"
#include <iostream>

int main(int argc, char* argv[]){

    int n, myid, numprocs, i;

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
    MPI_Comm_rank(MPI_COMM_WORLD, &myid);

    printf("Enter n: ");
    scanf("%d", &n);

    MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);

    if(myid==0){
        ++n;
    }

    if(myid==1){
        n+=2;
    }

    printf("n from process %d is %d\n", myid, n);

    return 0;
}

When executing the code this is the result (-np 3):

Enter n: Enter n: Enter n: 5

n from process 0 is 6

Every process execute the scanf function but it seems that only the 0 prints the results. I wanted to ask why it happens, although i can figure that this is a misuse of the parallel input.