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 <iostream>
#include <vector>
#include <queue>
#include<map>
#include<algorithm>
#include<set>
#define rep(i,n) for(int i = 0;i < n;i++)
#define rrep(i,n) for(int i = n-1;i >=0;i--)
#define ALL(a) a.begin(),a.end()
using namespace std;
int main() {
    int n,mx,mi; cin >> n; vector<int> a(n);
    rep(i, n) cin >> a[i];
    mx = *max_element(ALL(a)); mi = *min_element(ALL(a));
    cout << 2 * n - 1 << endl;
    if (abs(mx) >= abs(mi)) {
        int mxid = max_element(ALL(a)) - a.begin()+1;
        rep(i, n) {
            a[i] += mx;
            cout << mxid << " " << i+1 << endl;
        }
        rep(i, n-1)  cout << i + 1 << " " << i+2 << endl;
    }
    else {
        int miid = min_element(ALL(a)) - a.begin()+1;
        rep(i, n) {
            a[i] += mi;
            cout << miid << " " << i + 1 << endl;
        }
        rrep(i, n - 1) cout << i+2 << " " << i + 1 << endl;// 足す法、足されるほう
    }
}