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 <bits/stdc++.h>
#define maxn 200086

using namespace std;

int n;
vector<vector<int> > v; 
int a[maxn];
bool vis[maxn];
vector<pair<int, int> > ans;

void solve(vector<int> a, vector<int> b){
	if(a.size() == 1 && b.size() == 1) return;
	ans.push_back({a[0], b[0]});
	for(int i = 1;i < b.size();i++) ans.push_back({a[0], b[i]});
	for(int i = 1;i < a.size();i++) ans.push_back({b[0], a[i]});
	ans.push_back({a[0], b[0]});
}

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]) continue;
		int x = i;
		vector<int> w;
		while(!vis[x]){
			vis[x] = true;
			w.push_back(x);
			x = a[x];
		}
		v.push_back(w);
	}
	for(int i = 0;i + 1 < v.size();i += 2) solve(v[i], v[i + 1]);
	if(v.size() == 1){
		vector<int> a = v[0];
		ans.push_back({a[0], a[1]}), ans.push_back({a[1], a[2]}), swap(a[1], a[2]);
		for(int i = 3;i < a.size();i++) ans.push_back({a[2], a[i]});
		ans.push_back({a[0], a[1]}), ans.push_back({a[0], a[2]});
	}else if(v.size() & 1){
		vector<int> a;
		a.push_back(v[0][0]);
		solve(a, v.back());
	}
	printf("%d\n", ans.size());
	for(int i = 0;i < ans.size();i++) printf("%d %d\n", ans[i].first, ans[i].second);
}