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 int long long
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rrep(i,n) for(int (i)=((n)-1);(i)>=0;(i)--)
#define itn int
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
const long long INF = 1LL << 60;
const int MOD = 1000000007;
signed main(void){
    int n; cin>>n;
    vector <int> a(n); 
    rep(i,n) cin>>a[i];
    auto ima = max_element(all(a));
    auto imi = min_element(all(a));
    int ma = *ima, mi = *imi;
    bool pom;
    cout<<2*n-1<<endl;
    int x;
    if(abs(ma)>=abs(mi)) pom=true , x = (ima - a.begin()); else pom = false, x = imi - a.begin();
    rep(i,n) {if(pom) a[i]+= ma; else a[i]-=mi; cout<<x+1<<' '<<i+1<<endl;}
    //rep(i,n) cout<<a[i]<<endl;
    if(pom) rep(i,n-1) cout<<i+1<<' '<<i+2<<endl;
    else rrep(i,n-1) cout<<i+2<<' '<<i+1<<endl;
}
