B. Fixed Points

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears exactly once in it. For example, sequence [0, 2, 1] is a permutation of length 3 while both [0, 2, 2] and [1, 2, 3] are not.

A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer i is a fixed point of permutation a0, a1, ..., an - 1 if and only if ai = i. For example, permutation [0, 2, 1] has 1 fixed point and permutation [0, 1, 2] has 3 fixed points.

You are given permutation a. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.

Input

The first line contains a single integer n (1 ≤ n ≤ 105). The second line contains n integers a0, a1, ..., an - 1 — the given permutation.

Output

Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.

Sample test(s)

input

5
0 1 3 4 2

output

3

解题说明：此题是判断在一次交换之后【不一定是相邻交换】满足a[i]=i的数有多少。首先统计出交换之前a[i]=i的个数，然后遍历查找是否有两个数交换之后正好满足条件，如果找不到这样的两个数，最多只能再让一个数满足条件。当然，如果输入本来就是有序的，那就不需要查找了，直接输出n即可。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;



int main()
{
	int n,i,j,k,m,a[100001];
	scanf("%d",&n);
	m=0;
	k=0;
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		if (a[i]==i)
		{
			m++;
		}
	}
	for(i=0;i<n && !k;i++)
	{
		if (a[i]!=i && a[a[i]]==i)
		{
			k=1;
		}
	}
	printf("%d\n",m==n?n:m+k+1);
	return 0;
}