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 <cstdio>
#include <cctype>
#include <vector>
#include <cmath>

using namespace std;

const int MAXN = 400005;

int n, dest[MAXN];
bool vis[MAXN];

vector<pair<int, int>> ans;

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { (x *= 10) += ch - '0'; ch = getchar(); }
	return x * f;
}

inline void act(int x, int y)
{
	swap(dest[x], dest[y]);
	dest[x] = -dest[x], dest[y] = -dest[y];
	ans.push_back(make_pair(x, y));
}

inline void swap_circle(int x, int y)
{
	act(x, y);
	int p = x;
	while (dest[-dest[p]] > 0)
		act(p, -dest[p]);
	p = -dest[p];
	while (dest[-dest[p]] > 0)
		act(p, -dest[p]);
	act(p, -dest[p]);
}

int main()
{
	n = read();
	for (int i = 1; i <= n; i++)
		dest[i] = read();

	int oth = -1;
	for (int i = 1; i <= n; i++)
	{
		if (vis[i] || i == dest[i]) continue;
		int p = dest[i]; vis[i] = true;
		while (p != i)
			vis[p] = true, p = dest[p];
		if (oth == -1) oth = i;
		else swap_circle(i, oth), oth = -1;
	}

	if (oth != -1)
	{
		bool grand_circle = true;
		for (int i = 1; i <= n; i++)
			if (dest[i] == i)
			{ grand_circle = false, swap_circle(i, oth); break; }

		if (grand_circle)
		{
			int p = 1, tp = dest[p], ttp = dest[tp];
			act(p, tp); swap_circle(tp, ttp);
		}
	}

	printf("%d\n", (int)ans.size());
	for (pair<int, int> x : ans)
		printf("%d %d\n", x.first, x.second);

	return 0;
}