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:
//luogu AC!
#include <bits/stdc++.h>
using namespace std;

const int MAXN = 2e5 + 5;
typedef pair<int, int> pii;
int c[MAXN], vis[MAXN], re[MAXN];
int n, tot, tail;
vector<int> v[MAXN];
pii ans[MAXN];

inline void Swap(int x, int y) { ans[++tail] = { x,y };}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> c[i], re[c[i]] = i;
	for (int i = 1; i <= n; i++)
		if (c[i] != i && !vis[i]) {
			v[++tot].push_back(i);
			for (int j = c[i]; j != i; j = c[j])
				v[tot].push_back(j), vis[j] = 1;
		}
	for (int i = 2; i <= tot; i += 2) {
		Swap(v[i - 1][0], v[i][0]);
		for (int j = 1; j < v[i - 1].size(); j++)
			Swap(v[i - 1][j - 1], v[i - 1][j]);
		for (int j = 1; j < v[i].size(); j++)
			Swap(v[i][j - 1], v[i][j]);
		Swap(v[i - 1].back(), v[i].back());
	}
	if (tot & 1) {
		if (v[tot].size() == 2) {
			int pos = 1;
			while (pos == v[tot][0] || pos == v[tot][1]) pos++;
			Swap(v[tot][1], pos), Swap(v[tot][0], v[tot][1]), Swap(v[tot][0], pos);
		}
		else {
			Swap(v[tot][0], v[tot][1]);
			for (int i = 2; i < v[tot].size() - 1; i++)
				Swap(v[tot][i - 1], v[tot][i]);
			Swap(v[tot][0], v[tot].back()), Swap(v[tot][0], v[tot][v[tot].size() - 1 - 1]), Swap(v[tot][0], v[tot].back());
		}
	}
	cout << tail << '\n';
	for (int i = 1; i <= tail; i++)
		cout << re[ans[i].first] << ' ' << re[ans[i].second] << '\n', swap(re[ans[i].first], re[ans[i].second]);
}