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
// #define FILE_IN_OUT
#define RET(_x) cout << (_x) << '\n'; return;
#define all(_obj) (_obj).begin(), (_obj).end()
#define loop(_n) for (int i = 0; i < (_n); ++i)
#define sz(_obj) static_cast<int>((_obj).size())
#ifdef SHJ_LOCAL
#define debug(_x) std::cerr << (#_x) << " = " << (_x) << '\n' << std::flush;
#else
#define debug(_x) {}
#endif
#define endl "\n"
template <typename P, typename Q>
P CeilDiv(P _dividend, Q _divider) {
    return static_cast<P>((_dividend - 1LL + _divider) / _divider);
}
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
    for (Tp current = _begin; current != _end; ++current) {
        std::cout << *current << _delim;
    }
    std::cout << '\n';
}

//.constant
using ll = int64_t;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<int>(1e9 + 7);

//.data

//.code
void SolveTask() {
    int n;
    cin >> n;
    vector<int> arr(n);
    loop(n) {
        cin >> arr[i];
        --arr[i];
    }
    vector<vector<int>> cycles;
    vector<char> used(n, false);
    loop(n) {
        if (!used[i]) {
            cycles.emplace_back();
            for (int j = i; !used[j]; j = arr[j]) {
                used[j] = true;
                cycles.back().push_back(j);
            }
        }
    }
    vector<pii> ans;
    auto Swap = [&ans, &arr](int a, int b) {
        ans.emplace_back(a, b);
        swap(arr[a], arr[b]);
    };
    for (int i = 1; i < sz(cycles); i += 2) {
        int pos1 = cycles[i - 1][0];
        int pos2 = cycles[i][0];
        Swap(cycles[i - 1][0], cycles[i][0]);
        do {
            while (arr[pos1] != pos2) {
                Swap(pos1, arr[pos1]);
            }
            while (arr[pos2] != pos1) {
                Swap(pos2, arr[pos2]);
            }
        } while (arr[pos1] != pos2);
        Swap(pos1, pos2);
    }
    if ((sz(cycles) & 1) && sz(cycles.back()) > 1) {
        int a = cycles.back()[0];
        int b = arr[a];
        int c = arr[b];
        if (c == a) {
            loop(n) {
                if (i != a && i != b) {
                    c = i;
                    break;
                }
            }
            Swap(a, c);
            Swap(b, c);
            Swap(a, c);
        } else {
            Swap(a, b);
            Swap(b, c);
            Swap(c, a);
            while (arr[b] != a) {
                Swap(b, arr[b]);
            }
            Swap(a, b);
        }
    }
    cout << sz(ans) << endl;
    for (auto [a, b] : ans) {
        cout << a + 1 << ' ' << b + 1 << endl;
    }
}

int main() {
#ifdef FILE_IN_OUT
    std::ifstream cin("input.txt", std::ios::in);
    std::ofstream cout("output.txt", std::ios::out);
#else
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
#endif
    int test_cases = 1;
    // cin >> test_cases;
    for (int yt = 1; yt <= test_cases; ++yt) {
        SolveTask();
    }
#ifdef FILE_IN_OUT
    cin.close();
    cout.close();
#endif
    return 0;
}