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 For(a,b,c) for(int a=b;a<=c;++a)
#define Dor(a,b,c) for(int a=b;a>=c;--a)
#define CC(_...) fprintf(stderr,_)
using namespace std;
typedef long long LL;
typedef int Ni[200007];
int n,c,t;
Ni T,V,C,A,B,Z;
void O_O(int x,int y) {
	A[++t]=x,B[t]=y,Z[T[x]]^=1,Z[T[y]]^=1;
	swap(T[x],T[y]);
}
void X_X(int u) {
	while(!Z[T[u]]) u=T[u];
	while(!Z[T[T[u]]]) O_O(u,T[u]);
	u=T[u];
	while(Z[T[u]]) O_O(u,T[u]);
}
int main() {
	scanf("%d",&n);
	For(i,1,n) scanf("%d",T+i);
	For(i,1,n) if(!V[i]) {
		int u=i;
		while(!V[u]) V[u]=i,u=T[u];
		C[++c]=i;
	}
	for(int i=1;i<c;i+=2) {
		O_O(C[i],C[i+1]);
		X_X(C[i]);
	}
	if(c&1) {
		if(c>1) {
			For(i,1,n) if(C[c]!=i&&T[i]==i) {
				O_O(i,C[c]);
				break;
			}
			X_X(C[1]);
		}
		else {
			int x=T[1],y=T[x];
			O_O(1,x),O_O(y,x);
			X_X(1);
		}
	}
	printf("%d\n",t);
	For(i,1,t) printf("%d %d\n",A[i],B[i]);
	return 0;
}
  		   				 	  	   		  	 	  	 	