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)

#include "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }
///(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)///


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vi v(n);
	int maxv = -INF, minv = INF;
	int maxp = -1, minp = -1;
	REP(i, n) {
		cin >> v[i];
		if (maxv < v[i]) {
			maxv = v[i];
			maxp = i;
		}
		if (minv > v[i]) {
			minv = v[i];
			minp = i;
		}
	}
	int cnt = 0;
	
	vector<pii> op;
	if (abs(maxv) > abs(minv)||(maxv==minv&&maxv>0)) {
		assert(maxv >= 0);
		FOR(i, 1, n) {
			while (v[i] < v[i - 1]) {
				cnt++;
				v[i] += maxv;
				op.push_back(pii{ maxp + 1,i + 1 });
			}
			if (v[i] > maxv) {
				maxv = v[i];
				maxp = i;
			}

		}
	}
	else {
		assert(minv <= 0);
		REP(i, n - 1) {
			while (v[i+1] < v[i]) {
				cnt++;
				v[i] += minv;
				op.push_back(pii{ minp + 1,i + 1 });
			}
			if (v[i] < minv) {
				minv = v[i];
				minp = i;
			}
		}
	}
	cout << cnt << endl;
	REP(i, cnt) {
		cout << op[i].first << " " << op[i].second << endl;
	}
	

}