Byteazar 有 N 个小猪存钱罐. 每个存钱罐只能用钥匙打开或者砸开. Byteazar 已经把每个存钱罐的钥匙放到了某些存钱罐里. Byteazar 现在想买一台汽车于是要把所有的钱都取出来. 他想尽量少的打破存钱罐取出所有的钱,问最少要打破多少个存钱罐.

第一行一个整数 N (1 <= N <= 1.000.000) – 表示存钱罐的总数. 接下来每行一个整数,第 i+1行的整数代表第i个存钱罐的钥匙放置的存钱罐编号.

一个整数表示最少打破多少个存钱罐.

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <stack>

#include <vector>

#include <queue>

#include <cstring>

#include <string>

#include <map>

#include <set>

using namespace std;

const int BufferSize = 1 << 16;

char buffer[BufferSize], *Head, *Tail;

inline char Getchar() {

	if(Head == Tail) {

		int l = fread(buffer, 1, BufferSize, stdin);

		Tail = (Head = buffer) + l;

	}

	return *Head++;

}

int read() {

	int x = 0, f = 1; char c = Getchar();

	while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }

	while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }

	return x * f;

}

#define maxn 1000010

int n, fa[maxn], pos[maxn];

int findset(int x) { return x == fa[x] ? x : fa[x] = findset(fa[x]); }

int main() {

	n = read();

	for(int i = 1; i <= n; i++) fa[i] = i;

	for(int i = 1; i <= n; i++) {

		int u = findset(i), v = findset(read());

		if(u != v) fa[v] = u;

	}

	for(int i = 1; i <= n; i++) pos[i] = findset(i);

	sort(pos + 1, pos + n + 1);

	int cnt = 0;

	for(int i = 1; i <= n; i++) if(i == 1 || pos[i] != pos[i-1]) cnt++;

	printf("%d\n", cnt);

	return 0;

}