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;
typedef long long LL;

const LL inf  = 1e18+9;
const LL mod  = 1e9+7;
const LL maxn = 2e5+8;

LL n, m, c[maxn], vs[maxn], block[maxn], belong[maxn], sz[maxn], tot;

void process(vector<pair<LL, LL>> &vc, LL x, LL y) {
    LL ox = x, oy = y;
    swap(x, y);
    vc.push_back({x, y});
    while (c[x] != oy) vc.push_back({ox, c[x]}), x = c[x];
    while (c[y] != ox) vc.push_back({oy, c[y]}), y = c[y];
    vc.push_back({ox, oy});
}

void solve() {
    LL i, j, x, y;
    scanf("%lld", &n);
    vector<pair<LL, LL>> vc;
    for (i = 1; i <= n; i++) scanf("%lld", &c[i]);
    for (i = 1; i <= n; i++) {
        if (c[i] == i) vs[i] = 1;
        else if (!vs[i]) {
            block[++tot] =  j = i;
            do {
                sz[i]++;
                belong[j] = i; vs[j] = 1; j = c[j];
            } while (j != i);
        }
    }
    if (tot & 1) {
        x = block[tot];
        if (sz[block[tot]] == 2) {
            for (y = 1; belong[y] == x; y++);
            vc.push_back({x, y});
            vc.push_back({y, c[x]});
            vc.push_back({x, y});
        }
        else {
            y = c[x];
            LL ox = x, oy = y;
            vc.push_back({x, y});
            swap(x, y);
            while (c[c[x]] != ox) vc.push_back({ox, c[x]}), x = c[x];
            vc.push_back({oy, c[x]});
            vc.push_back({ox, c[x]});
            vc.push_back({ox, oy});
        }
    }
    for (i = 1; i < tot; i += 2) process(vc, block[i], block[i + 1]);
    printf("%lld\n", (LL)vc.size());
    for (auto it : vc) printf("%lld %lld\n", it.first, it.second);
}

signed main() {
    LL t = 1; //scanf("%lld", &t);
    while (t--) solve();
    return 0;
}