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;
typedef double db;
typedef long long ll;
typedef pair<int, int> pii;
#define fi first
#define se second
#define empb emplace_back
#define all(x) (x).begin(),(x).end()

int main() {
#ifdef local
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	vector<int> a(n + 1);
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	vector<vector<int> > cycles;
	vector<int> vis(n + 1);
	for(int i = 1; i <= n; i++) {
		if(vis[i]) {
			continue;
		}
		vector<int> cyc;
		int t = i;
		while(!vis[t]) {
			vis[t] = 1;
			cyc.push_back(t);
			t = a[t];
		}
		if(cyc.size() != 1) cycles.push_back(cyc);
	}
	int sz = cycles.size();
	vector<pii> ans;
	for(int i = 1; i < sz; i += 2) {
		auto c1 = cycles[i];
		auto c2 = cycles[i - 1];
		ans.emplace_back(c1[0], c2[0]);
		for(int j = 1; j < (int) c1.size(); j++) {
			ans.emplace_back(c1[j], c2[0]);
		}
		for(int j = 1; j < (int) c2.size(); j++) {
			ans.emplace_back(c2[j], c1[0]);
		}
		ans.emplace_back(c1[0], c2[0]);
	}
	if(sz & 1) {
		auto c = cycles.back();
		if(c.size() > 2) {
			ans.emplace_back(c[0], c[1]);
			for(int i = 2; i < (int) c.size(); i++) {
				ans.emplace_back(c[1], c[i]);
			}
			ans.emplace_back(c[0], c[2]);
			ans.emplace_back(c[0], c[1]);
		} else if(c.size() == 2) {
			int t = 1;
			while(t == c[0] || t == c[1]) t++;
			ans.emplace_back(c[0], t);
			ans.emplace_back(c[1], t);
			ans.emplace_back(c[0], t);
		}
	}
	cout << ans.size() << '\n';
	for(auto &e : ans) {
		cout << e.fi << ' ' << e.se << '\n';
		swap(a[e.fi], a[e.se]);
		a[e.fi] = -a[e.fi];
		a[e.se] = -a[e.se];
	}
//	for(int i = 1; i <= n; i++) cout << a[i] << ' '; cout << endl;
	for(int i = 1; i <= n; i++) {
		assert(a[i] == i);
	}
	return 0;
}