#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int a[55];

int main()
{
	int N;
	cin >> N;
	cin >> a[0];
	P M = P(a[0],0), m = P(a[0],0);
	for (int i = 1; i < N; ++i)
	{
		cin >> a[i];
		if (a[i] > M.first) M = P(a[i],i);
		if (a[i] < m.first) m = P(a[i],i);
	}
	cout << 2*N - 2 << endl;
	if (abs(m.first) < abs(M.first))
	{
		rep(i,N)
		{
			if (i != M.second) cout << M.second+1 << " " << i+1 << endl;
		}
		rep(i,N-1)
		{
			cout << i+1 << " " << i+2 << endl;
		}
	}
	else
	{
		rep(i,N)
		{
			if (i != m.second) cout << m.second+1 << " " << i+1 << endl;
		}
		rep(i,N-1)
		{
			cout << N-i << " " << N-i-1 << endl;
		}
	}
	return 0;
}
### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int a[55];

int main()
{
	int N;
	cin >> N;
	cin >> a[0];
	P M = P(a[0],0), m = P(a[0],0);
	for (int i = 1; i < N; ++i)
	{
		cin >> a[i];
		if (a[i] > M.first) M = P(a[i],i);
		if (a[i] < m.first) m = P(a[i],i);
	}
	cout << 2*N - 2 << endl;
	if (abs(m.first) < abs(M.first))
	{
		rep(i,N)
		{
			if (i != M.second) cout << M.second+1 << " " << i+1 << endl;
		}
		rep(i,N-1)
		{
			cout << i+1 << " " << i+2 << endl;
		}
	}
	else
	{
		rep(i,N)
		{
			if (i != m.second) cout << m.second+1 << " " << i+1 << endl;
		}
		rep(i,N-1)
		{
			cout << N-i << " " << N-i-1 << endl;
		}
	}
	return 0;
}
```