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>
#define For(i,a,b) for(register int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(register int i=(a);i>=(b);--i)
using namespace std;
inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;
#define maxn 200005
#define inf 0x3f3f3f3f

int n,a[maxn];
bool bo[maxn];
vector<pii>res;
inline void work(int x,int y){
	bo[x]^=1,bo[y]^=1;
	res.pb(mkp(x,y));
}
int id[maxn];
vector<int>v[maxn],o;

signed main()
{
	n=read();
	For(i,1,n)a[i]=read();
	For(i,1,n)
		if(!id[i]){
			id[i]=i;
			v[i].pb(i),o.pb(i);
			for(int t=a[i];t!=i;t=a[t])id[t]=i,v[i].pb(t);
		}
	for(int i=0;i+1<o.size();i+=2)
	{
		int x=o[i],y=o[i+1];
		work(v[x][0],v[y][0]);
		for(auto t:v[x])if(t!=v[x][0])work(v[y][0],t);
		for(auto t:v[y])if(t!=v[y][0])work(v[x][0],t);
		work(v[x][0],v[y][0]);
	}
	if(o.size()&1)
	{
		int x=o.back();
		if(v[x].size()!=1)
		{
			if(v[x].size()==2){
				int y=1;
				while(y==v[x][0]||y==v[x][1])++y;
				work(v[x][0],y);
				work(y,v[x][1]);
				work(v[x][0],y); 
			}
			else{
	//			int tmp=v[x].back();
	//			for(int i=(int)v[x].size()-1;i>=1;--i)
	//				v[x][i]=v[x][i-1];
	//			v[x][0]=tmp;
	//			work(v[x][0],v[x][2]);
	//			work(v[x][0],v[x][1]);
	//			swap(v[x][2],v[x][1]);
	//			work(v[x][0],v[x][1]);
//				for(int i=3;i<v[x].size();++i)
//					work(v[x][2],v[x][i]);
//				work(v[x][0],v[x][2]);
				
				for(int i=1;i<(int)v[x].size()-1;++i)
					work(v[x][0],v[x][i]);
				work(v[x][1],v[x].back());
				work(v[x][0],v[x].back());
				work(v[x][1],v[x][0]);
			}
		}
	}
	cout<<res.size()<<endl;
	for(auto t:res)cout<<t.fi<<' '<<t.se<<endl;
	return 0;
}