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>
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priority_queue
#define P pair<int,int>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const ll LINF=1e18;
const ld PI=abs(acos(-1));
int n,a[55];

int main(){
	cin >> n;
	REP(i,0,n) cin >> a[i]; int ma=0;
	REP(i,0,n) if(abs(a[i])>abs(a[ma])) ma=i;
	
	cout << 2*n-1 << endl;
	REP(i,0,n) cout << ma+1 << " " << i+1 << endl;
	if(a[ma]>0){
		REP(i,1,n) cout << i << " " << i+1 << endl;
	}else{
		RREP(i,n,2) cout << i << " " << i-1 << endl;
	}
	return 0;
}
