Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
#define repi(i,a,n) for(int i=a;i<n;i++)
#define mp make_pair
#define fastIO ios::sync_with_stdio(false)
#define read freopen("in.in","r",stdin)
#define write freopen("out.out","w",stdout)
#define scf(xx) scanf("%d",&xx)
#define scfl(xx) scanf("%I64d",&xx)
#define All(a) a.begin(), a.end()


typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, ll> ii;
typedef long double ld;



vector<ii> sol;
bool solve1(vi v){
	sol.clear();
	set<ii> st;
	rep(i, v.size())st.insert(mp(v[i], i));
	for (int i = 1; i < v.size(); i++){
		if (v[i] >= v[i - 1])continue;
		if (v[i] + st.rbegin()->first <= v[i])
			return false;
		while (v[i] < v[i - 1]){
			st.erase(mp(v[i], i));
			v[i] += st.rbegin()->first;
			sol.push_back(mp(st.rbegin()->second + 1, i + 1));
			st.insert(mp(v[i], i));
		}
	}
	return true;
}

bool solve2(vi v){
	sol.clear();
	set<ii> st;
	rep(i, v.size())st.insert(mp(v[i], i));
	for (int i = v.size() - 2;i >= 0; i--){
		if (v[i] <= v[i + 1])continue;
		if (v[i] + st.begin()->first >= v[i])
			return false;
		while (v[i] > v[i + 1]){
			st.erase(mp(v[i], i));
			v[i] += st.begin()->first;
			sol.push_back(mp(st.begin()->second + 1, i + 1));
			st.insert(mp(v[i], i));
		}
	}
	return true;
}
int main(){	
	fastIO;
	//write; read;
	int n;
	vi v;
	cin >> n;
	v.resize(n);
	rep(i, n)cin >> v[i];
	if (solve1(v) == false)
		solve2(v);
	cout << sol.size() << endl;
	rep(i, sol.size())
		cout << sol[i].first << " " << sol[i].second << endl;
	return 0;
}