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>
using namespace std;
#define int long long
#define pi pair<int,int>
#define f first
#define s second
const int nax=2e5+100;
int a[nax],vis[nax],n;
vector<pi> res;
void check(int i,int j)
{
    swap(a[i],a[j]);
    a[i]*=-1;
    a[j]*=-1;
    res.push_back({i,j});
}
void find(int i,int j)
{
    check(i,j);
    int cur=i;
    while(a[-a[cur]]>0)
    check(cur,-a[cur]);
    cur=-a[cur];
    while(a[-a[cur]]>0)
    check(cur,-a[cur]);
    check(cur,-a[cur]);
}
void solve()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    int p=-1;
    for(int i=1;i<=n;i++)
    if(!vis[i])
    {
        if(a[i]==i)
        continue;
        int cur=i;
        do
        {
            vis[cur]=1;
            cur=a[cur];
        }while(cur!=i);
        if(p==-1)
        p=i;
        else
        {
            find(p,i);
            p=-1;
        }
    }
    if(p!=-1)
    {
        bool f=0;
        for(int i=1;i<=n;i++)
        if(a[i]==i)
        {
            find(p,i);
            f=1;
            break;
        }
        if(!f)
        {
            int t1=a[p],t2=a[a[p]];
            check(p,t1);
            find(t1,t2);
        }
    }
    cout<<res.size()<<'\n';
    for(auto x:res)
    cout<<x.f<<' '<<x.s<<'\n';
}
int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    int t=1;
    //cin>>t;
    while(t--)
    solve();
    return 0;
}