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 FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define INF 100000000
typedef long long ll;

using namespace std;


int main()
{
    int n;
    int a[100];
    cin >> n;
    int imax = 0;
    int number = 0;
    REP(i, n) {
        cin >> a[i];
        if (imax < abs(a[i])) {
            number = i;
            imax = abs(a[i]);
        }
    }
    cout << 2*n-1 << endl;
    REP(i, n) {
        cout << number + 1 << " " << i + 1 << endl;
        a[i] += a[number];
    }
    if (a[number] >= 0) {
        REP(i, n - 1) {
            cout << i + 1 << " " << i + 2 << endl;
            a[i + 1] += a[i];
        }
    }
    else {
        for(int i=n-2;i>=0;i--) {
            cout << i + 2 << " " << i + 1 << endl;
            a[i] += a[i+1];
        }
    }
    return 0;
}