codeforces347B

Fixed Points

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, ..., a_n - 1 if and only if a_i = 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 nintegers a₀, a₁, ..., a_n - 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.

Examples

Input

5
0 1 3 4 2

Output

3

sol：容易发现交换最多使得答案增加2，一般都能加1（似乎并没什么用）
于是O(n)扫一遍能加2就加2，否则试试能不能加1

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=100005;
int n,a[N],Pos[N];
int main()
{
    int i,ans=0;
    R(n);
    for(i=1;i<=n;i++)
    {
        a[i]=(read()+1); Pos[a[i]]=i;
        if(a[i]==i) ans++;
    }
    for(i=1;i<=n;i++) if(a[i]!=i)
    {
        if(Pos[Pos[a[i]]]==a[i])
        {
            Wl(ans+2);
            return 0;
        }
    }
    Wl(min(ans+1,n));
    return 0;
}
/*
input
5
0 1 3 4 2
output
3
*/

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