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 endl '\n'
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define LL long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n,a[200010];
int m,vst[200010];
vector<int>c[200010];
vector<pii>ans;
void update(int x,int y){
	ans.emplace_back(x,y);
	swap(a[x],a[y]);
	a[x]=-a[x];
	a[y]=-a[y];
}
void solve(int A,int B){
	update(c[A][0],c[B][0]);
	while(a[-a[c[A][0]]]>0)update(c[A][0],-a[c[A][0]]);
	while(a[-a[c[B][0]]]>0)update(c[B][0],-a[c[B][0]]);
	update(c[A][0],c[B][0]);
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=n;i++){
		if(vst[i]||a[i]==i)continue;
		int p=i;
		c[++m].pb(i),vst[i]=1;
		while(a[p]!=i){
			p=a[p];
//			cout<<p<<endl;
//			cout<<endl<<endl<<endl;
			c[m].pb(p);
			vst[p]=1;
		}
	}
	for(int i=1;i+1<=m;i+=2)solve(i,i+1);
	if(m%2==1){
		if(c[m].size()==2){
			int pos=0;
			for(int i=1;i<=n;i++)if(i!=c[m][0]&&i!=c[m][1])pos=i;
			update(c[m][0],pos),update(c[m][1],pos),update(c[m][0],pos);
		}else if(c[m].size()>2){
			update(c[m][0],c[m].back()),update(c[m][0],c[m][1]);
			while(a[-a[c[m][0]]]>0)update(c[m][0],-a[c[m][0]]);
			while(a[-a[c[m].back()]]>0)update(c[m].back(),-a[c[m].back()]);
			update(c[m][0],c[m].back());
		}
	}
	cout<<ans.size()<<endl;
	for(auto i:ans)cout<<i.F<<' '<<i.S<<endl;
	return 0;
}