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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define DUMP(x) cout << #x << " = " << (x) << endl
#define MP make_pair
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef pair<int, int> PII;

int main() {
  int n; cin >> n;
  VI a(n);
  REP(i, n) cin >> a[i];
  int mx = (int)(max_element(a.begin(), a.end()) - a.begin());
  int mi = (int)(min_element(a.begin(), a.end()) - a.begin());
  vector<PII> op;
  int ans = 0;
  if (a[mi] < 0) {
    if (a[mx] >= a[mi]) {
      REP(i, n) {
        a[i] += a[mx];
        ans++;
        op.push_back(MP(mx + 1, i + 1));
      }
    } else {
      REP(i, n) {
        a[i] += a[mi];
        ans++;
        op.push_back(MP(mi + 1, i + 1));
      }
    }
  }
  FOR(i, 1, n) {
    a[i] += a[i - 1];
    ans++;
    op.push_back(MP(i, i + 1));
  }
  cout << ans << endl;
  REP(i, op.size()) {
    cout << op[i].first << " " << op[i].second << endl;
  }
  return 0;
}