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>
using namespace std;
int qabs(int a){
	return (a < 0) ? -a : a;
}

int n;
int p[200200];
int cnt = 0;
int bel[200200];
int fst_app[200200];
vector< pair<int, int> > ans;

void do_op(int a, int b){
	ans.push_back(make_pair(a, b));
	swap(p[a], p[b]);
	p[a] = -p[a], p[b] = -p[b];
}

int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> p[i];
	memset(bel, -1, sizeof(bel));
	for(int i = 1; i <= n; i++){
		if(bel[i] < 0){
			int now = i;
			fst_app[cnt] = i;
			do{
				bel[now] = cnt;
				now = p[now];
			}while(now != i);
			cnt++;
		}
	}
	if(cnt == 1){
		int a = qabs(p[1]), b = qabs(p[a]);
		do_op(1, a);
		do_op(a, b);
		int now = 1;
		while(p[qabs(p[now])] > 0) do_op(now, qabs(p[now]));
		now = a;
		while(p[qabs(p[now])] > 0) do_op(now, qabs(p[now]));
		do_op(1, now);
	} else {
		for(int i = 0; i + 1 < cnt; i += 2){
			do_op(fst_app[i], fst_app[i + 1]);
			int now = fst_app[i];
			while(p[qabs(p[now])] > 0) do_op(now, qabs(p[now]));
			now = fst_app[i + 1];
			while(p[qabs(p[now])] > 0) do_op(now, qabs(p[now]));
			do_op(fst_app[i], fst_app[i + 1]);
		}
		if(cnt % 2 == 1){
			do_op(fst_app[cnt - 1], 1);
			while(p[qabs(p[1])] > 0) do_op(1, qabs(p[1]));
			do_op(fst_app[cnt - 1], 1);
		}
	}
	cout << (int)ans.size() << endl;
	for(int i = 0; i < (int)ans.size(); i++){
		cout << ans[i].first << " " << ans[i].second << endl; 
	}
	return 0;
}