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.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(x) cout<<#x<<':'<<x<<endl
const int N = 2e5+20;
int cnt,f[N],a[N],col[N],vis[N];
int n,pos[N],tmp1,tmp2,tmp3;
vector<int> cir[N];
vector<pair<int,int> > ans;
void solve(int i,int j)
{
	ans.push_back(make_pair(i,j));
}
int main()
{
	scanf("%d",&n);
	for(int i=1,x;i<=n;i++)
		scanf("%d",&x),pos[i]=x;
	cnt = 0;
	for(int i=1;i<=n;i++)if(!vis[i])
	{
		vis[i] = 1;++cnt;
		cir[cnt].push_back(i);
		for(int j=pos[i];j!=i;j=pos[j])
			vis[j]=1,cir[cnt].push_back(j);
	}
	for(int i=1;i<=cnt;i+=2)
	{
		if(i+1>cnt)break;
		solve(cir[i][0],cir[i+1][0]);
		for(int j=1;j<cir[i].size();j++)
			solve(cir[i][j],cir[i+1][0]);
		for(int j=1;j<cir[i+1].size();j++)
			solve(cir[i+1][j],cir[i][0]);
		solve(cir[i][0],cir[i+1][0]);
	}
	if(cnt%2)
	{
		if(cir[cnt].size()==2)
		{
			int k = 1;
			while(k==cir[cnt][0]||k==cir[cnt][1])k++;
			solve(cir[cnt][0],k);solve(cir[cnt][1],k);solve(cir[cnt][0],k);
		}
		else if(cir[cnt].size()!=1)
		{
			for(int i=1;i<cir[cnt].size()-1;i++)solve(cir[cnt][0],cir[cnt][i]);
			solve(cir[cnt][1],cir[cnt].back());
			solve(cir[cnt][0],cir[cnt].back());
			solve(cir[cnt][0],cir[cnt][1]);
		}
	}
	printf("%d\n",ans.size());
	for(auto x:ans)
		printf("%d %d\n",x.first,x.second);
	return 0;
}
