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.
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
const int N=200100;
int a[N],vis[N];
int n;
vector<int>cyc;
vector<pair<int,int>>res;

void dfs(int idx){
	vis[idx]=true;
	cyc.push_back(idx);
	if(!vis[a[idx]])dfs(a[idx]);
}

int mex(vector<int>&c){
	int cur=0;
	while(cur<c.size()&&c[cur]==cur)++cur;
	return cur;
}

int solve2(vector<int>c1,vector<int>c2);

int solve1(vector<int>c){
	if(c.size()==1)return 1;
	if(c.size()==2)return solve2(c,{mex(c)});
	int n=c.size();
	res.push_back({c[n-2],c[n-1]});
	res.push_back({c[n-3],c[n-1]});
	for(int i=0;i<n-2;++i)res.push_back({c[i],c[n-2]});
	res.push_back({c[n-2],c[n-1]});
	return 1;
}

int solve2(vector<int>c1,vector<int>c2){
	if(c1.size()==1)return solve1(c2);
	for(auto p:c1)res.push_back({p,c2.back()});
	for(auto p:c2)res.push_back({c1[0],p});
	return 1;
}
 
int main(){
	cin.tie(0);ios::sync_with_stdio(0);
	cin>>n;
	for(int i=0;i<n;++i)cin>>a[i];
	for(int i=0;i<n;++i)--a[i];
	vector<vector<int>>cycles;
	for(int i=0;i<n;++i)if(!vis[i]){
		dfs(i);
		cycles.push_back(cyc);
		cyc.clear();
	}
	for(int i=0;i<cycles.size();i+=2){
		if(i+1<cycles.size())solve2(cycles[i],cycles[i+1]);
		else solve1(cycles[i]);
	}
	cout<<res.size()<<endl;
	for(auto r:res)cout<<r.first+1<<" "<<r.second+1<<endl;
	assert(res.size()<=n+1);
}
