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;

const int N = 2e5 + 3;

int n, a[N];
bool vis[N];
vector<array<int, 2>> ins;

void get_cycle(int u, vector<int> &cycle){
    cycle.push_back(u);
    vis[u] = true;

    if(!vis[a[u]]) get_cycle(a[u], cycle);
}

void do_2_cycles(const vector<int> &c1, const vector<int> &c2){
    for(int i = 0; i < c2.size(); ++i)
        ins.push_back({c1[0], c2[i]});
    for(int i = 1; i < c1.size(); ++i)
        ins.push_back({c1[i], c2[0]});
    ins.push_back({c1[0], c2[0]});
}

void do_1_cycle(const vector<int> &cycle){
    if(cycle.size() == 2){
        int add;
        for(int i = 1; i <= n; ++i)
            if(cycle[0] != i && cycle[1] != i){
                add = i;
                break;
            }

        ins.push_back({cycle[0], add});
        ins.push_back({cycle[1], add});
        ins.push_back({cycle[0], add});
        return;
    }

    for(int i = 1; i < (int)cycle.size() - 1; ++i)
        ins.push_back({cycle[0], cycle[i]});
    ins.push_back({cycle[1], cycle.back()});
    ins.push_back({cycle[0], cycle.back()});
    ins.push_back({cycle[0], cycle[1]});
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];

    vector<int> prev_cycle;
    for(int i = 1; i <= n; ++i)
        if(!vis[i]){
            vector<int> cycle;
            get_cycle(i, cycle);
            if(cycle.size() == 1) continue;
            if(prev_cycle.empty()){
                prev_cycle = cycle;
                continue;
            }
            do_2_cycles(cycle, prev_cycle);
            prev_cycle = {};
        }

    if(prev_cycle.size()) do_1_cycle(prev_cycle);

    cout << ins.size() << "\n";
    for(auto [x, y]: ins)
        cout << x << " " << y << "\n";
}