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<map>
#include<set>
#include<string>
#include<utility>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<queue>
#include<stack>

#define ll int64_t
#define Rep(i, n) for (ll i = 0; i < n; i++)

using namespace std;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;
    vector<ll> a(N, 0);
    ll far = 0;
    Rep (i, N) {
        cin >> a[i];
        if (abs(a[i]) > abs(a[far])) {
            far = i;
        }
    }

    ll m = 2 * N - 1;
    cout << m << "\n";
    Rep (i, N) {
        cout << far+1 << " " << i+1 << "\n";
    }

    if (a[far] > 0) {
        Rep (i, N-1) {
            cout << i+1 << " " << i+2 << "\n";
        }
    } else {
        Rep (i, N-1) {
            cout << N-i << " " << N-i-1 << "\n";
        }
    }

    cout << flush;
}