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>

using namespace std;

#define rep(i, a) for (int i = 0; i < a; i++)
#define reps(i, a, b) for (int i = a; i < b; i++)

int n;
long long a[50];

void culcp() {
	rep(i, n-1) {
		cout << i+1 << " " << i+2 << endl;
	}
}

void culcm() {
	rep(i, n-1) {
		cout << n-i << " " << n-i-1 << endl;
	}
}

void plusma(int mai) {
	rep(i, n) {
		cout << mai << " " << i+1 << endl;
	}
	culcp();
}
	
void plusmi(int mii) {
	rep(i, n) {
		cout << mii << " " << i+1 << endl;
	}
	culcm();
}

int main()
{
	cin >> n;
	rep(i, n) cin >> a[i];
	long long mi, ma;
	int mai, mii;
	mi = ma = a[0];
	mai = mii = 1;
	rep(i, n) {
		if (mi > a[i]) mi = a[i], mii = i+1;
		if (ma < a[i]) ma = a[i], mai = i+1;
	}
	if (mi >= 0) {
		cout << n-1 << endl;
		culcp();
	} else if (ma <= 0) {
		cout << n-1 << endl;
		culcm();
	} else if (ma > -mi) {
		cout << 2*n-1 << endl;
		plusma(mai);
	} else {
		cout << 2*n-1 << endl;
		plusmi(mii);
	}
	return 0;
}