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;
const int maxn=2e5;
int n;
int c[maxn];
vector<pair<int,int>> res;
void cswap(int pos1,int pos2){
	res.push_back({pos1,pos2});
	swap(c[pos1],c[pos2]);
	c[pos1]=~c[pos1];
	c[pos2]=~c[pos2];
}
void solve(){
	cin >> n;
	for(int i=0;i<n;i++){
		cin >> c[i];
		c[i]--;
	}
	vector<vector<int>> cycle;
	vector<bool> vis(n,false);
	for(int i=0;i<n;i++){
		if(vis[i]) continue;
		vector<int> cyc;
		int j=i;
		while(!vis[j]){
			cyc.push_back(j);
			vis[j]=true;
			j=c[j];
		}
		cycle.push_back(cyc);
	}
	while(cycle.size()>=2){
		vector<int> cyc1=cycle.back();
		cycle.pop_back();
		vector<int> cyc2=cycle.back();
		cycle.pop_back();
		for(int i=0;i<cyc1.size();i++)
			cswap(cyc1[i],cyc2[0]);
		for(int i=1;i<cyc2.size();i++)
			cswap(cyc1[0],cyc2[i]);
		cswap(cyc1[0],cyc2[0]);
	}
	if(cycle.size()){
		vector<int> cyc=cycle[0];
		if(cyc.size()==2){
			int other=c[0] ? (c[1] ? 1 : 2) : 0;
			cswap(cyc[0],other);
			cswap(cyc[1],other);
			cswap(cyc[0],other);
		}
		if(cyc.size()>2){
			cswap(cyc[0],cyc[1]);
			for(int i=2;i<cyc.size();i++)
				cswap(cyc[1],cyc[i]);
			cswap(cyc[0],cyc[2]);
			cswap(cyc[0],cyc[1]);
		}
	}
	cout << res.size() << endl;
	for(pair<int,int> operation: res)
		cout << operation.first+1 << ' ' << operation.second+1 << endl;
}
int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	int t=1;
	while(t--)
		solve();
	return 0;
}