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 f first
#define s second
using namespace std;
typedef long long ll;
const int N = 1e3+5, mod = 1e9+7;
int n, a[N];
vector<pair<int, int>> ans;
void solvePOS()
{
	for(int i=1; i<n; i++) ans.push_back({i, i+1});
}
void solveNEG()
{
	for(int i=n-1; i>0; i--) ans.push_back({i+1, i});
}
void print()
{
	cout << ans.size() << endl;
	for(auto i : ans) cout << i.f << " " << i.s << endl;
}
int main()
{
	cin >> n;
	for(int i=0; i<n; i++) cin >> a[i];
	int mn = *min_element(a, a+n);
	int mx = *max_element(a, a+n);
	if(mn >= 0) solvePOS();
	else if(mx <= 0) solveNEG();
	else if(mx >= abs(mn))
	{
		int l = find(a, a+n, mx)-a + 1;
		for(int i=0; i<n; i++) if(a[i] < 0) ans.push_back({l, i+1});
		solvePOS();
	}
	else
	{
		int l = find(a, a+n, mn)-a + 1;
		for(int i=0; i<n; i++) if(a[i] > 0) ans.push_back({l, i+1});
		solveNEG();
	}
	print();
}