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;

#define mp make_pair
#define fr first
#define sc second

long long n,nn=0,a[200069],pst[200069],ex[200069],zs=0;
bitset<200069> vtd;
pair<long long,long long> sq[200069];

int main()
{
	long long i,ii,k,p;
	
	scanf("%lld",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&k);
		a[k]=i;
		pst[i]=k;
	}
	for(i=1;i<=n;i++)
	{
		if(!vtd[i])
		{
			nn++;
			ex[nn]=i;
			for(p=i;!vtd[p];p=a[p])
			{
				vtd[p]=1;
			}
		}
	}
	if(nn>1)
	{
		if(nn%2)
		{
			zs++;
			sq[zs]={a[ex[nn]],a[ex[1]]};
			for(p=pst[ex[nn]];p!=ex[nn];p=pst[p])
			{
				zs++;
				sq[zs]={a[p],a[ex[1]]};
			}
			zs++;
			sq[zs]={a[ex[nn]],a[ex[1]]};
		}
		for(i=1;i<nn;i+=2)
		{
			zs++;
			sq[zs]={a[ex[i]],a[ex[i+1]]};
			for(ii=0;ii<2;ii++)
			{
				for(p=pst[ex[i+ii]];p!=ex[i+ii];p=pst[p])
				{
					zs++;
					sq[zs]={a[p],a[ex[i+!ii]]};
				}
			}
			zs++;
			sq[zs]={a[ex[i]],a[ex[i+1]]};
		}
	}
	else
	{
		zs++;
		sq[zs]={a[1],1};
		zs++;
		sq[zs]={1,pst[1]};
		zs++;
		sq[zs]={pst[1],a[1]};
		for(p=pst[pst[pst[1]]];p!=1;p=pst[p])
		{
			zs++;
			sq[zs]={a[p],1};
		}
		zs++;
		sq[zs]={a[1],1};
	}
	printf("%lld\n",zs);
	for(i=1;i<=zs;i++)
	{
		printf("%lld %lld\n",sq[i].fr,sq[i].sc);
	}
}