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.
import sys
input = sys.stdin.readline

n = int(input())
A = [0] + list(map(int, input().split()))
seen = [0] * (n + 1)
cycles = []
ans = []
for i in range(1, n + 1):
    cur = []
    j = i
    while not seen[j]:
        cur.append(j)
        seen[j] = 1
        j = A[j]
    if len(cur) > 1: cycles.append(cur)

for i in range(1, len(cycles), 2):
    X, Y = cycles[i - 1], cycles[i]
    ans.append((X[0], Y[0]))
    for x in X[1:]: ans.append((x, Y[0]))
    for y in Y[1:]: ans.append((y, X[0]))
    ans.append((X[0], Y[0]))
if len(cycles) % 2:
    X = cycles[-1]
    if len(X) > 2:
        for x in X[:-2]: ans.append((x, X[-1]))
        ans.append((X[0], X[-2]))
        ans.append((X[-2], X[-1]))
        ans.append((X[0], X[-1]))
    else:
        Y = []
        for i in range(1, n + 1):
            if i != X[0] and i != X[1]: break
        ans.append((i, X[0]))
        ans.append((i, X[1]))
        ans.append((i, X[0]))
print(len(ans))
for a in ans: print(*a)