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<cstdio>
#include<vector>
#define pii std::pair<int,int>
#define mp std::make_pair
#define N 200005

int n,a[N];

bool vis[N];

std::vector<std::vector<int> > c;

std::vector<pii> ans;

inline void sol(std::vector<int> &x,std::vector<int> &y){
//	printf("&%d %d\n",x.size(),y.size());
	int u=x[0],v=y[0];
	ans.push_back(mp(u,v));
	for(auto i:x)
		if(i!=u)
			ans.push_back(mp(v,i));
	for(auto i:y)
		if(i!=v)
			ans.push_back(mp(u,i));
	ans.push_back(mp(u,v));
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		if(!vis[i]){
			std::vector<int> tmp;
			tmp.clear();
			for(int j=i;!vis[j];j=a[j]){
				vis[j]=1;
				tmp.push_back(j);
			}
			if(tmp.size()>1)
				c.push_back(tmp);
		}
	if(c.size()==1&&c[0].size()==n){
		ans.push_back(mp(c[0][0],c[0][1]));
		ans.push_back(mp(c[0][1],c[0][2]));
		ans.push_back(mp(c[0][0],c[0][2]));
		for(int i=4;i<=n;i++)
			ans.push_back(mp(c[0][1],c[0][i-1]));
		ans.push_back(mp(c[0][0],c[0][1]));
	}
	else{
		while(c.size()>1){
			std::vector<int> x=c.back();
			c.pop_back();
			std::vector<int> y=c.back();
			c.pop_back();
			sol(x,y);
		}
		if(c.size()){
			std::vector<int> x=c.back(),y;
			y.clear();
			for(auto i:x)
				vis[i]=0;
			for(int i=1;i<=n;i++)
				if(vis[i]){
					y.push_back(i);
					break;
				}
			sol(x,y);
		}
	}
	printf("%d\n",ans.size());
	for(auto i:ans)
		printf("%d %d\n",i.first,i.second);
}