#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>
#include <sstream>
#include <utility>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#define CLOSE() ios::sync_with_stdio(false)
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define IN() freopen("in.txt", "r", stdin)
#define OUT() freopen("out.txt", "w", stdout)
const int maxn = 100005;
using LL = long long;
using UI = unsigned int;
using namespace std;
//------------------------------------------------------------------------------------------//
//暴力法，乘法速度慢，用i*i超时，sqrt能过，貌似sqrt是用牛顿迭代还是二分实现的？
int main() {
int n;
while (scanf("%d", &n) == 1) {
int cnt = 0;
while (n--) {
int x;
scanf("%d", &x);
int flag = 1;
//注意sqrt写外面
int t = sqrt(x)+1;
for (int j = 2; j < t; ++j) {
if (x % j == 0) {
flag = 0;
break;
}
}
if (flag) cnt++;
}
printf("%d\n", cnt);
}
return 0;
}
//Miller_Rabin算法
//费马小定理：假设p是质数，那么对任意a，a^(p-1)=1(mod p)。
//利用这一定理，只要对任意取s次a属于[1, p-1]，mod = 1，则p为素数。
//二次探测定理：p为素数，x方 = 1（mod p）解为x = 1 || x = p-1.
//把p-1拆分成为二进制表示2^t*u，令x[0] = a^u，随后平方t次，若出现
//某次x[i] mod p = 1但是x[i-1] != 1 && x[i-1] != p-1,则p非素数。
//若数据大，则将乘法也用快速幂的方法表示。
const int S = 4;
LL q_pow(LL a, LL n, LL mod) {
LL ret = 1;
while (n) {
if (n & 1) ret = ret*a%mod;
n >>= 1;
a = a*a%mod;
}
return ret;
}
bool Miller_Rabin(LL n) {
if (n == 2) return true;
if (n < 2 || !(n & 1)) return false;
int t = 0;
LL a, x, y, u = n - 1;
while ((u & 1) == 0) t++, u >>= 1;
for (int i = 0; i < S; i++) {
a = rand() % (n - 1) + 1;
x = q_pow(a, u, n);
for (int j = 0; j < t; j++) {
y = x*x%n;
if (y == 1 && x != 1 && x != n - 1)
return false;
x = y;
}
if (x != 1) return false;
}
return true;
}
int main() {
int n;
while (scanf("%d", &n) == 1) {
int x, cnt = 0;
while (n--) {
scanf("%d", &x);
if (Miller_Rabin(x)) cnt++;
}
printf("%d\n", cnt);
}
return 0;
}
