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 int long long

int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    int n;
    cin >> n;
    int a[n],d,pos[n];
    for(int i=0;i<n;i++){
    	cin >> a[i];
    	--a[i];
    	pos[a[i]]=i;
    }
    vector <vector <int> > cy;
    vector <int> vis(n,0);
    for(int i=0;i<n;i++){
    	if(vis[i])continue;
    	vector <int> c;
    	int x=i;
    	while(!vis[x]){
    		c.push_back(x);
    		vis[x]=1;
    		x=a[x];
    	}
    	if(c.size()>1)cy.push_back(c);
    	else d=c[0];
    }
    vector <pair <int,int> > ans;
    for(int i=0;i+1<cy.size();i+=2){
    	int x=cy[i].back(),y=cy[i+1].back();
    	ans.push_back({x,y});
    	for(int j=0;j+1<cy[i].size();j++)
    		ans.push_back({y,cy[i][j]});
    	for(int j=0;j+1<cy[i+1].size();j++)
    		ans.push_back({x,cy[i+1][j]});
    	ans.push_back({x,y});
    	d=cy[i][0];
    }
    if(cy.size()%2){
    	auto c=cy.back();
    	if(c.size()==2){
    		ans.push_back({d,c[0]});
    		ans.push_back({d,c[1]});
    		ans.push_back({d,c[0]});
    	}
    	else{
    		int l=c.size();
    		ans.push_back({c[l-2],c[l-3]});
    		ans.push_back({c[l-2],c[l-1]});
    		ans.push_back({c[l-3],c[l-1]});
    		for(int i=0;i<l-2;i++)
    			ans.push_back({c[i],c[l-2]});
    	}
    }
    cout << ans.size() << '\n';
    for(auto [x,y]:ans)
    	cout << x+1 << ' ' << y+1 << '\n';
}