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 <bits/stdc++.h>
#define maxn 55
using namespace std;
typedef long long LL;
int a[maxn];
int main() {
//    freopen("in.txt", "r", stdin);
    int n;
    scanf("%d", &n);
    printf("%d\n", 2*n-2);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    int p1=1, p2=1;
    for (int i = 2; i <= n; ++i) {
        if (a[i] > a[p1]) p1 = i;
        if (a[i] < a[p2]) p2 = i;
    }
    if (abs(a[p1]) >= abs(a[p2])) {
        for (int i = 1; i < p1; ++i) printf("%d %d\n", p1, i);
        for (int i = p1+1; i <= n; ++i) printf("%d %d\n", p1, i);
        for (int i = 2; i <= n; ++i) printf("%d %d\n", i-1, i);
    }
    else {
        for (int i = 1; i < p2; ++i) printf("%d %d\n", p2, i);
        for (int i = p2+1; i <= n; ++i) printf("%d %d\n", p2, i);
        for (int i = n-1; i >= 1; --i) printf("%d %d\n", i+1, i);
    }
    return 0;
}
