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 <iostream>
#include <vector>
#include <set>
#include <cassert>

using namespace std;

int main() {
    int n; cin >> n;
    vector<int> c(1+n);
    for (int i = 1; i <= n; ++i) cin >> c[i];
    vector< pair<int, int> > ops; // operations

    vector<bool> seen(1+n);
    for (int i = 1; i <= n; ++i) {
        if (c[i] == i) seen[i] = true;
    }
    int waiting = 0;

    auto swp = [&](int pos1, int pos2) {
        ops.emplace_back(pos1, pos2);
        swap(c[pos1], c[pos2]);
    };

    // fix 2 loops a and b in size_a + size_b operations
    auto pairLoop = [&](int a, int b) {
        swp(a, b);
        while (c[a] != b) swp(a, c[a]);
        while (c[b] != a) swp(b, c[b]);
        swp(a, b);
    };

    for (int i = 1; i <= n; ++i) {
        if (seen[i]) continue;
        int node = i;
        do {
            seen[node] = true;
            node = c[node];
        } while (node != i);
        if (!waiting) waiting = i;
        else {
            pairLoop(i, waiting);
            waiting = 0;
        }
    }

    if (waiting) {
        int inPlace = 0;
        for (int i = 1; i <= n; ++i) {
            if (c[i] == i) {
                inPlace = i;
                break;
            }
        }
        if (inPlace) {
            pairLoop(inPlace, waiting);
        } else { // All nodes in one loop, thus at least 3 nodes
            // cancel 1 loop a in size_a + 1 operations
            assert(c[c[waiting]] != waiting);
            int x = c[waiting];
            swp(waiting, x);
            while (c[c[waiting]] != waiting) swp(waiting, c[waiting]);

            int y = waiting, z = c[waiting]; // x and z are flipped coins, y is not
            swp(x, z);
            swp(y, z);
            swp(x, y);
        }
    }

    for (int i = 1; i <= n; ++i) assert(c[i] == i);
    assert(ops.size() <= n+1);
    cout << ops.size() << endl;
    for (pair<int, int> o : ops) cout << o.first << ' ' << o.second << endl;
    return 0;
}