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>

const int N = 200000;

int c[N + 5], n; bool b[N + 5];

std::vector<std::pair<int, int> >ans;
void print() {
	printf("%d\n", (int)ans.size());
	for(auto x : ans) printf("%d %d\n", x.first, x.second);
}
void add(int x, int y) {
	ans.push_back(std::make_pair(x, y));
	b[c[x]] ^= 1, b[c[y]] ^= 1;
	std::swap(c[x], c[y]);
}

void do1(int x, int y) {
	add(x, y);
	while( c[y] != x ) add(y, c[y]);
	while( c[x] != y ) add(x, c[x]);
	add(x, y);
}
void do2(int x) {
	if( c[x] == x ) return ;
	int y = c[x]; add(x, y);
	while( c[c[x]] != x ) add(x, c[x]);
	int z = c[x];
	add(y, z), add(x, z), add(x, y);
}
void do3(int x, int y, int z) {
	add(x, y), add(y, z);
	while( c[y] != x ) add(y, c[y]);
	while( c[x] != y ) add(x, c[x]);
	add(x, y);	
}

int siz[N + 5]; bool vis[N + 5]; std::vector<int>v;
int main() {
	scanf("%d", &n); for(int i=1;i<=n;i++) scanf("%d", &c[i]);
	for(int i=1;i<=n;i++) if( !vis[i] ) {
		v.push_back(i);
		for(int p=i;!vis[p];p=c[p])
			vis[p] = true;
	}
	
	if( v.size() == 1 ) do2(v[0]);
	else if( v.size() & 1 ) {
		for(unsigned i=3;i+1<v.size();i+=2) do1(v[i], v[i + 1]);
		do3(v[0], v[1], v[2]);
	} else {
		for(unsigned i=0;i+1<v.size();i+=2) do1(v[i], v[i + 1]);
	}
	
	if( (int)ans.size() > n + 1 ) {puts("WA 1"); return 0;}
	for(int i=1;i<=n;i++) {
		if( b[i] ) {puts("WA 2"); return 0;}
		if( c[i] != i ) {puts("WA 3"); return 0;}
	}
	print();
}