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>
#include <complex>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define fi first
#define se second
#define mp make_pair
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B,C,L,R;
string S,T;
const ll MOD = (1e+9) + 7;
const ll INF = 1LL << 60;
typedef pair<ll,ll> P;

typedef vector<P> vp;
typedef vector<vp> matP;


int main() {
    cin>>N;
    vec a(N);
    rep(i,N) cin>>a[i];
    ll ma, mi;
    rep(i,N){
        if(a[i] == *max_element(a.begin(), a.end())) ma = i;
        if(a[i] == *min_element(a.begin(), a.end())) mi = i;
    }
    cout<<N*2 - 1<<endl;
    if(a[ma] > abs(a[mi])){
        rep(i,N) cout<<ma+1<<' '<<i+1<<endl;
        rep(i,N-1) cout<<i+1<<' '<<i+2<<endl;
    }else{
        rep(i,N) cout<<mi+1<<' '<<i+1<<endl;
        rep(i,N-1) cout<<N-i<<' '<<N-1-i<<endl;
    }
}