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 N 200020
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],tot,vis[N];
vector<pair<int,int> > ans;
vector<int> g[N];
inline void Swap(int x,int y){
	ans.emplace_back(x,y);
	swap(a[x],a[y]);
	a[x]*=-1,a[y]*=-1;
}
void Solve1(vector<int> &A,vector<int> &B){
	Swap(A[0],B[0]);
	for(int i=1;i<(int)A.size();++i)Swap(A[i],B[0]);
	for(int i=1;i<(int)B.size();++i)Swap(B[i],A[0]);
	Swap(A[0],B[0]);
}
void Solve2(vector<int> &g){
	int sz=g.size();
	for(int i=2;i<sz;++i){
		Swap(g[0],g[i]);
	}
	Swap(g[0],g[1]);
	Swap(g[1],g[2]);
	Swap(g[0],g[2]);
}
int main(){
	n=read();
	for(int i=1;i<=n;++i)a[i]=read();
	for(int i=1;i<=n;++i){
		if(a[i]==i)continue;
		if(vis[i])continue;
		++tot;
		for(int u=i;;u=a[u]){
			g[tot].push_back(u);
			vis[u]=1;
			if(a[u]==i)break;
		}
	}
	for(int i=1;i+1<=tot;i+=2){
		Solve1(g[i],g[i+1]);
	}
	if(tot&1){
		if(g[tot].size()==2){
			int p=1;
			while(p==g[tot][0]||p==g[tot][1])++p;
			Swap(p,g[tot][0]),Swap(p,g[tot][1]),Swap(p,g[tot][0]);
		}
		else{
			Solve2(g[tot]);
		}
	}
	printf("%d\n",(int)ans.size());
	for(auto [x,y]:ans){
		printf("%d %d\n",x,y);
	}
	return 0;
}