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;

#define sz(x) ((int) (x).size())
typedef long long ll;
typedef long double ld;

int n, c[200100], t[200100];
bool u[200100];
vector<pair<int, int>> r;

void sw(int i, int j) {
  t[c[i]]++, t[c[j]]++;
  r.push_back({i, j});
  swap(c[i], c[j]);
}

void solve(int i) {
  while (c[i] != i)
    sw(i, c[i]);
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(12);
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> c[i], c[i]--;
  int x = -1;
  for (int i = 0; i < n; i++) {
    if (c[i] == i || u[i])
      continue;
    int j = i;
    do {
      u[j] = true;
      j = c[j];
    } while (j != i);
    if (x == -1)
      x = i;
    else {
      sw(x, i);
      sw(c[x], c[i]);
      solve(c[x]);
      solve(c[i]);
      x = -1;
    }
  }
  if (x != -1) {
    if (c[c[x]] == x) {
      int i = 0;
      while (i == x || i == c[x])
        i++;
      sw(i, c[x]);
      sw(i, x);
      sw(i, c[i]);
    } else {
      int i = x, j = c[x];
      while (c[c[i]] != i)
        sw(i, c[i]);
      sw(j, c[i]);
      sw(i, c[i]);
      sw(i, c[i]);
    }
  }
  // cout << "Fin\n";
  // for (int i = 0; i < n; i++)
  //   cout << c[i] << (i < n - 1 ? " " : "\n");
  // for (int i = 0; i < n; i++)
  //   cout << t[i] << (i < n - 1 ? " " : "\n");
  cout << sz(r) << "\n";
  for (auto & p : r)
    cout << p.first + 1 << " " << p.second + 1 << "\n";
}