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 <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

void Swap(const int p1, const int p2, vector<pair<int, int>> &ans) {
    int p3 = 0;
    while (p3 == p1 || p3 == p2) ++p3;
    ans.push_back(make_pair(p2, p3));
    ans.push_back(make_pair(p1, p3));
    ans.push_back(make_pair(p2, p3));
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        --a[i];
    }

    vector<vector<int>> pp;
    vector<int> w(n, 0);
    for (int i = 0; i < n; ++i) {
        if (i == a[i] || w[i]) {
            continue;
        }
        vector<int> p;
        for (int j = i; !w[j]; j = a[j]) {
            w[j] = 1;
            p.push_back(j);
        }
        pp.push_back(p);
    }

    vector<pair<int,int>> ans;
    while (pp.size() > 1) {
        vector<int> p1 = pp.back();
        pp.pop_back();
        vector<int> p2 = pp.back();
        pp.pop_back();
        ans.push_back(make_pair(p1[0], p2[0]));
        for (int i = 1; i < p2.size(); ++i) {
            ans.push_back(make_pair(p1[0], p2[i]));
        }
        for (int i = 1; i < p1.size(); ++i) {
            ans.push_back(make_pair(p1[i], p2[0]));
        }
        ans.push_back(make_pair(p1[0], p2[0]));
    }

    if (!pp.empty()) {
        vector<int> p = pp.back();
        if (p.size() == 2) {
            Swap(p[0], p[1], ans);
        }
        else {
            for (int i = 0; i + 2 < (int)p.size(); ++i) {
                ans.push_back(make_pair(p[i], p.back()));
            }
            int p1 = p[0];
            int p3 = p.back();
            int p2 = p[p.size()-2];
            ans.push_back(make_pair(p1, p2));
            ans.push_back(make_pair(p2, p3));
            ans.push_back(make_pair(p1, p3));
        }
    }

    cout << ans.size() << endl;
    for (int i = 0; i < (int)ans.size(); ++i) {
        cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
    }
 
    return 0;
}