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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
int n,a[1000005],m,tans[1000005][2],t,vis[1000005];
vector<int> val[200005];
void getswap(int x,int y)
{
	swap(a[x],a[y]);
	tans[++t][0]=x;
	tans[t][1]=y;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		if(a[i]!=i&&!vis[i])
		{
			m++;
			int nw=i;
			while(!vis[nw])
			{
				val[m].push_back(nw);
				vis[nw]=1;
				nw=a[nw];
			}
		}
	for(int i=1;i+1<=m;i+=2)
	{
		int sz0=val[i].size(),sz1=val[i+1].size();
		int u=val[i][0],v=val[i+1][0];
		getswap(u,v);
		for(int j=1;j<sz1;j++)
			getswap(u,val[i+1][j]);
		for(int j=1;j<sz0;j++)
			getswap(v,val[i][j]);
		getswap(u,v);
	}
	if(m%2)
	{
		int sz=val[m].size();
		if(sz==2)
		{
			int u=val[m][0],v=val[m][1],id=1;
			while(id==u||id==v) id++;
			getswap(u,id);
			getswap(v,id);
			getswap(u,id);
		}
		else
		{
			int u=val[m][0],v=val[m][1],w=val[m][2];
			getswap(u,v);
			getswap(v,w);
			getswap(u,w);
			for(int i=3;i<sz;i++)
				getswap(v,val[m][i]);
			getswap(u,v);
		}
	}
	for(int i=1;i<=n;i++)
		fprintf(stderr,"%d ",a[i]);
	fprintf(stderr,"\n");
	printf("%d\n",t);
	for(int i=1;i<=t;i++)
		printf("%d %d\n",tans[i][0],tans[i][1]);
	return 0;
}