Snuke has an integer sequence, a, of length N. The i-th element of a (1-indexed) is a_{i}.

He can perform the following operation any number of times:

* Operation: Choose integers x and y between 1 and N (inclusive), and add a_x to a_y.



He would like to perform this operation between 0 and 2N times (inclusive) so that a satisfies the condition below. Show one such sequence of operations. It can be proved that such a sequence of operations always exists under the constraints in this problem.

* Condition: a_1 \leq a_2 \leq ... \leq a_{N}

Constraints

* 2 \leq N \leq 50
* -10^{6} \leq a_i \leq 10^{6}
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_{N}


Output

Let m be the number of operations in your solution. In the first line, print m. In the i-th of the subsequent m lines, print the numbers x and y chosen in the i-th operation, with a space in between. The output will be considered correct if m is between 0 and 2N (inclusive) and a satisfies the condition after the m operations.

Examples

Input

3
-2 5 -1


Output

2
2 3
3 3


Input

2
-1 -3


Output

1
2 1


Input

5
0 0 0 0 0


Output

0
# -*- coding: utf-8 -*-
import sys
N=int(sys.stdin.readline().strip())
A=map(int, sys.stdin.readline().split())
A_max=max(A)
A_max_idx=A.index(A_max)    #0-indexed
A_min=min(A)
A_min_idx=A.index(A_min)    #0-indexed

if 0<=A_min:    #Aの最小値が0以上である場合、累積和を構成すれば良い
    print N-1
    for i in range(1,N):
        print i,i+1
else:    #Aの最小値が0より小さい場合
    if 0<=A_max:
        if abs(A_min)<=abs(A_max):  #絶対を取った時にA_maxがA_min以上
            print 2*N-1
            for i in range(1,N+1):  #A_maxを配列Aの全ての値にそれぞれ足していく
                print A_max_idx+1,i
            for i in range(1,N):    #その後に累積和を取れば良い
                print i,i+1
        elif abs(A_min)>abs(A_max):  #絶対を取った時にA_minがA_max以上
            print 2*N-1
            for i in range(1,N+1):  #A_minを配列Aの全ての値にそれぞれ足していって、aを全て負の数にする
                print A_min_idx+1,i
            for i in range(N,1,-1): #後ろから累積和
                print i,i-1
    elif A_max<0:   #aの値が全部負なので、後ろから累積和を取ればよい
        print N-1
        for i in range(N,1,-1): #後ろから累積和
            print i,i-1