347_B. Fixed Points_34933
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.

Examples

Input

5
0 1 3 4 2


Output

3
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.StringTokenizer;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.floor;
import static java.lang.Math.ceil;
import static java.lang.Math.pow;
import static java.lang.Math.log;
import static java.lang.Math.sqrt;

/* ------------ Theorems and Lemmas ------------ */

// Fermat's Little Theorem
// (1) ((a ** p) - a) % p = 0; if 'p' is prime
// (2) Further, if 'a % p != 0' (a and b should be coprime i.e. gcd(a, b) = 1), then ((a ** (p - 1)) - 1) % p = 0

/* -------------- End of theorems -------------- */

public class Solution {

	// Digit DP
	private static int digit;
	private static int times;
	private static StringBuilder number;
	private static int[][][] digitDP;
	// End

	protected static final String[] FACTS = new String[] { "", "", "2", "3", "322", "5", "53", "7", "7222", "7332" };

	public static void main(String[] args) {
		FastReader reader = new FastReader();
		int t = 1;
		// t = reader.nextInt();
		StringBuilder ans = new StringBuilder("");
		outer: while (t-- > 0) {
			int n = reader.nextInt();
			int cnt = 0;
			int[] a = new int[n];
			for (int i = 0; i < n; ++i)
				cnt += (a[i] = reader.nextInt()) == i ? 1 : 0;
			if (cnt >= n - 1) {
				ans.append(cnt + "\n");
				continue;
			}
			for (int i = 0; i < n; ++i)
				if (a[i] != i && a[a[i]] == i) {
					ans.append(cnt + 2 + "\n");
					continue outer;
				}
			ans.append(cnt + 1 + "\n");
		}
		reader.close();
		System.out.println(ans.toString().trim());
	}

	public static boolean prime(int n) {
		if (n <= 1)
			return false;
		if (n == 2)
			return true;
		if ((n & 1) == 0)
			return false;
		int val = (int) sqrt((double) n);
		for (int i = 3; i <= val; ++i)
			if (n % i == 0)
				return false;
		return true;
	}

	public static int ddp(int n) {
		number = new StringBuilder(String.valueOf(n));
		digitDP = new int[12][12][2];
		for (int[][] a : digitDP)
			for (int[] b : a)
				Arrays.fill(b, -1);
		return call(0, 0, 0);
	}

	// Digit DP
	public static int call(int pos, int cnt, int f) {
		if (cnt > times)
			return 0;
		if (pos == number.length())
			return cnt == times ? 1 : 0;
		if (digitDP[pos][cnt][f] != -1)
			return digitDP[pos][cnt][f];
		int res = 0;
		int lmt = f == 0 ? number.charAt(pos) - '0' : 9;
		for (int dgt = 0; dgt <= lmt; ++dgt)
			if ((dgt == digit ? 1 + cnt : cnt) <= times)
				res += call(1 + pos, dgt == digit ? 1 + cnt : cnt, f == 0 && dgt < lmt ? 1 : f);
		digitDP[pos][cnt][f] = res;
		return res;
	}

	// Binary Exponentiation
	public static int binpow(int a, int n, int mod) {
		// Iterative approach
		a %= mod;
		int result = 1;
		while (n > 0) {
			if ((n & 1) != 0)
				result = result * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return result;

		// Recursive approach
		/*
		 * if (n == 0l) return 1l; long res = binpow(a, n / 2, mod); if ((n & 1l) != 0l)
		 * return ((res % mod) * (res % mod) * (a % mod)) % mod; else return ((res %
		 * mod) * (res % mod)) % mod;
		 */
	}

	// GCD
	public static int gcd(int a, int b) {
		while (b > 0) {
			a %= b;
			int t = a;
			a = b;
			b = t;
		}
		return a;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine().trim());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] readArray(int size) {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = nextInt();
			return ar;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}
}