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
#include <bits/stdc++.h>
using namespace std;
#define rp(i,s,e) for(int i=(s);i<(e);++i)
#define irp(i,s,e) for(int i=(s);i>(e);--i)

int main(){
	int N; cin>>N;
	int minA = INT_MAX, maxA = INT_MIN;
	int minIx, maxIx;
	rp(i, 1, N+1){
		int a; cin>>a;
		if(maxA < a){ maxA = a; maxIx = i; }
		if(minA > a){ minA = a; minIx = i; }
	}

	if(minA == maxA) puts("0");//same
	else if(minA>=0 && maxA>=0){//all +
		cout<< N-1 <<endl;
		rp(i,1,N) cout<<i<<" "<<i+1<<endl;
	}
	else if(minA<=0 && maxA<=0){//all -
		cout<< N-1 <<endl;
		irp(i,N,1) cout<<i<<" "<<i-1<<endl;
	}
	else if(minA<=0 && maxA>=0){//mix -+
		cout<< 2*N-1 <<endl;
		if(abs(minA) > abs(maxA)){//|min|
			rp(i,1,N+1) cout<<minIx<<" "<<i<<endl;
			irp(i,N,1) cout<<i<<" "<<i-1<<endl;
		}
		// else if(abs(minA) > abs(maxA)){
		else{//|max|
			rp(i,1,N+1) cout<<maxIx<<" "<<i<<endl;
			rp(i,1,N) cout<<i<<" "<<i+1<<endl;
		}
	}

}
