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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ld = long double;
const int INF = 1 << 29;
const ll LINF = 1l << 61;
const ld PI = abs(acos(-1));
#define pq priotity_queue
#define mp make_pair
#define P pair
#define rep(i, a, b) for (ll i = a; i < b; i++)
#define rrep(i, a, b) for (ll i = a; i >= b; i--)
#define cbuf(buf, init) memset(buf, init, sizeof(buf))
#define modp(x, y) (((y) + (x) % (y)) % (y))
#define divd(x, y) (static_cast<double>(x) / static_cast<double>(y))

ll N, a[100], b[100];

int main()
{
    //
    cin >> N;
    rep(i, 0, N) { cin >> a[i]; }

    rep(i, 0, N - 1) { b[i] = a[i] - a[i + 1]; }

    ll cnt = 0;
    vector<pair<int, int>> ans;
    while (true) {
        ll mi = max_element(b, b + N - 1) - b;
        if (b[mi] <= 0) break;
        ll besti = 0, bestj = 0, bestv = LINF;
        bool left = false;
        rep(i, 0, N - 1)
        {
            if (b[i] <= 0) continue;
            rep(j, 0, N)
            {
                if (abs(a[i] + a[j] - a[i + 1]) < bestv) {
                    bestv = a[i] + a[j] - a[i + 1];
                    bestj = j;
                    left = true;
                    besti = i;
                }
                if (abs(a[i] - (a[i + 1] + a[j])) < bestv) {
                    bestv = a[i] - (a[i + 1] + a[j]);
                    bestj = j;
                    left = false;
                    besti = i;
                }
            }
        }
        if (left) {
            a[besti] += a[bestj];
            ans.push_back(mp(besti, bestj));
        }
        else {
            a[besti + 1] += a[bestj];
            ans.push_back(mp(besti + 1, bestj));
        }

        /*
        ll Mi = max_element(a, a + N) - a;
        ll M = a[Mi];
        ll mi = min_element(a, a + N) - a;
        ll m = a[mi];
        if (a[i] + m - a[i + 1] < a[i] - (a[i + 1] + M)) {
            a[i] += m;
            ans.push_back(mp(i, mi));
        }
        else {
            a[i + 1] += M;
            ans.push_back(mp(i + 1, Mi));
        }
        */
        rep(i, 0, N - 1) { b[i] = a[i] - a[i + 1]; }
        cnt++;
        if (cnt > 2 * N) throw std::runtime_error("");
    }
    cout << ans.size() << endl;
    for (auto p : ans) cout << p.second + 1 << " " << p.first + 1 << endl;
}