//离散数学中轮换与对换的关系 轮换长为n的可以换成n-1个对换

include[HTML_REMOVED]

using namespace std;

const int maxn = 1e4 + 10;
int vis[maxn],map[maxn];
int n, cnt;

int main()
{
cin >> n;
for(int i = 1; i <= n;i) cin >> map[i];
for(int i = 1; i <= n;i)
{
int x = map[i],t = x;
if(x != i && !vis[i])
{
vis[x] = 1;
x = map[x];

        while(x != t)
        {
            vis[x] = 1;
            x = map[x];
            cnt ++;
        }

    }
}
cout << cnt;
return 0;

}