#include <bits/stdc++.h>
#define rep(i,n) for(ll i=0;i<(n);i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;
template <typename T>
using vec = std::vector<T>;

int main() {
    int N;
    cin >> N;
    vec<int> a(N);
    int mx = 0;
    int mn = 0;
    rep(i, N) {
        cin >> a[i];
        mx = max(mx, a[i]);
        mn = min(mn, a[i]);
    }

    cout << N*2-1 << endl;
    if(mx >= abs(mn)){
        int idx = find(all(a),mx) - a.begin() + 1;
        rep(i,N) cout << idx << " " << i+1 << endl;
        rep(i,N-1) cout << i+1 << " " << i+2 << endl;
    } else {
        int idx = find(all(a),mn) - a.begin() + 1;
        rep(i,N) cout << idx << " " << i+1 << endl;
        for(int i=N;i>=2;--i) cout << i << " " << i-1 << endl;
    }
}
### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for(ll i=0;i<(n);i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;
template <typename T>
using vec = std::vector<T>;

int main() {
    int N;
    cin >> N;
    vec<int> a(N);
    int mx = 0;
    int mn = 0;
    rep(i, N) {
        cin >> a[i];
        mx = max(mx, a[i]);
        mn = min(mn, a[i]);
    }

    cout << N*2-1 << endl;
    if(mx >= abs(mn)){
        int idx = find(all(a),mx) - a.begin() + 1;
        rep(i,N) cout << idx << " " << i+1 << endl;
        rep(i,N-1) cout << i+1 << " " << i+2 << endl;
    } else {
        int idx = find(all(a),mn) - a.begin() + 1;
        rep(i,N) cout << idx << " " << i+1 << endl;
        for(int i=N;i>=2;--i) cout << i << " " << i-1 << endl;
    }
}
```