#include<fstream.h>
int p[170]={
		0,1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,
		47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,
		113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,
		197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,
		281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,
		379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,
		463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,
		571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,
		659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,
		761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,
		863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,
		977,983,991,997};

int i,j,n,k;
unsigned long int a[100][100];
unsigned long int sum = 0;
int v[100];

//a[i][j]  cate siruri se termina cu j si au divizor p[i]

int main(){
  ifstream f("indep.in");
  f>>n;
  for (i=1;i<=n;i++)
    f>>v[i];
  f.close();
  for (i=2;i<=169;i++){
    if (v[1]<p[i]) break;

    if (v[1]%p[i]==0)
      a[i][1]=1;
  }

  for (j=2;j<=n;j++){
    for (i=2;i<=169;i++){
      if (v[j]<p[i]) break;
      if (v[j]%p[i]!=0) break;
      a[i][j]=1;
      for (k=1;k<j;k++)
	a[i][j]+=a[i][k];
    }
  }
  for (j=1;j<=n;j++){
    for (i=1;i<=169;i++)
      if (p[i]>v[j]) break;
      else sum+=a[i][j];
  }
  ofstream g("indep.out");
  g<< ((1<<n)-sum)<<endl;
  g.close();
  return 0;
}