Problem: There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins. 

In one operation, you can do the following:

  * Choose 2 distinct indices i and j.

  * Then, swap the coins on positions i and j.

  * Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)




Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.

Note that you do not need to minimize the number of operations.

Input

The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.

The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).

Output

In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.

In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.

Examples

Input


3
2 1 3


Output


3
1 3
3 2
3 1


Input


5
1 2 3 4 5


Output


0

Note

Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.

The series of moves performed in the first sample changes the coins as such:

  * [~~~2,~~~1,~~~3] 
  * [-3,~~~1,-2] 
  * [-3,~~~2,-1] 
  * [~~~1,~~~2,~~~3] 



In the second sample, the coins are already in their correct positions so there is no need to swap.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < int(n); i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb push_back

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 200005;
int n, a[MX], vis[MX];
vvi v;
vii res;

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	
	for (int i = 1; i <= n; i++)
		if (a[i] != i && !vis[i]) {
			vi w;
			int u = i;
			
			do {
				vis[u] = 1;
				w.pb(u);
				u = a[u];
			} while (u != i);
			
			v.pb(w);
		}
	
	for (int i = 0; i + 1 < (int)v.size(); i += 2) {
		vi &a = v[i];
		vi &b = v[i + 1];
		
		res.emplace_back(a.back(), b.back());
		
		for (int i = 0; i + 1 < (int)a.size(); i++)
			res.emplace_back(b.back(), a[i]);
			
		for (int i = 0; i + 1 < (int)b.size(); i++)
			res.emplace_back(a.back(), b[i]);
		
		res.emplace_back(a.back(), b.back());
	}
	
	if (v.size() & 1) {
		vi &a = v.back();
		int a3;
		
		if (a.size() == 2) {
			int a3 = 1;
			while (a3 == a[0] || a3 == a[1])
				a3++;
			res.emplace_back(a[0], a3);
			res.emplace_back(a[1], a3);
			res.emplace_back(a[0], a3);
		} else {
			for (int i = 1; i + 1 < (int)a.size(); i++)
				res.emplace_back(a[0], a[i]);
			res.emplace_back(a[1], a.back());
			res.emplace_back(a[0], a.back());
			res.emplace_back(a[0], a[1]);
		}
	}
	
	cout << res.size() << endl;
	for (auto &[a, b] : res)
		cout << a << " " << b << endl;
	
	return 0;
}