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;
int n,c[200005],tot,X[200005],Y[200005];
bool vis[200005];
vector<int>v1,v2;
void get(int x,vector<int>&v)
{
    int now=x;
    do
    {
        v.push_back(now);
        vis[now]=1;
        now=c[now];
    }
    while(now!=x);
}
void sw(int x,int y){tot++;X[tot]=x;Y[tot]=y;swap(c[x],c[y]);}
void work()
{
    int t1=v1.size(),t2=v2.size();
    sw(v1[0],v2[0]);
    for(int i=1;i<t1;i++)sw(v1[i],v2[0]);
    for(int i=1;i<t2;i++)sw(v2[i],v1[0]);
    sw(v1[0],v2[0]);
    v1.clear(),v2.clear();
}
void work2()
{
    if(v1.size()==2)
    {
        for(int i=1;i<=n;i++)if(v1[0]!=i&&v1[1]!=i){v1.push_back(i);break;}
        sw(v1[0],v1[2]);
        sw(v1[1],v1[2]);
        sw(v1[0],v1[2]);
        return;
    }
    sw(v1[0],v1[1]);
    for(int i=2;i<v1.size()-1;i++)sw(v1[i],v1[0]);
    sw(v1[1],v1[v1.size()-1]);
    sw(v1[0],v1[v1.size()-1]);
    sw(v1[0],v1[1]);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&c[i]);
    for(int i=1;i<=n;i++)
    {
        if(c[i]==i||vis[i])continue;
        if(!v1.size())get(i,v1);
        else
        {
            get(i,v2);
            work();
        }
    }
    if(v1.size())work2();
    printf("%d\n",tot);
    for(int i=1;i<=tot;i++)printf("%d %d\n",X[i],Y[i]);
    return 0;
}