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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <cmath>
#include <iomanip>
#include <functional>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP(i, n) for (int i = 1; i < (int)(n); ++i)
using namespace std;
using lint = int64_t;

int main() {
    int N;
    cin >> N;
    vector<int> a(N);
    rep(i, N) cin >> a[i];
    
    cout << 2 * (N - 1) << "\n";
    vector<pair<int, int>> S(N);
    rep(i, N) S[i] = pair<int, int>{abs(a[i]), i};
    sort(S.rbegin(), S.rend());
    rep(i, N) {
        if (i == S[0].second) continue;
        cout << S[0].second + 1 << " " << i + 1 << "\n";
    }
    if (a[S[0].second] > 0) {
        rep(i, N - 1) cout << i + 1 << " " << i + 2 << "\n";
    } else {
        for (int i = N - 1; i >= 1; --i) cout << i + 1 << " " << i << "\n";
    }
    return 0;
}
