
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve(ll k)
{
ll i;
for ( i = 2; i <= sqrt(k); i++)
{
if (k % i == 0)
{
cout << "noprime";
break;
}
}
if (i > sqrt(k))
cout << "prime";
}
int main()
{
ll n;
cin >> n;
ll a, b=1;
string n1 = to_string(n); //to_string函数可以将数字变为字符串
n /= 10;
while (b)
{
b = n / 10;
a = n % 10;
string z = to_string(a);
n1 += z;
n = b;
}
ll n2 = atoll(n1.c_str()); //atoll函数是将字符串变为长长整型
solve(n2);
return 0;
}