对于这种构造问题
1. 如果想要构造出单调不递减我们可以假设把它都构造同成同一个值
2. 接下来该如何思考呢?我们会发现这两种操作从奇偶性并没有发现任何性质,但根据l,r两个索引顺序我们可以从两边观察a[1]与a[n]是否相等,将两个数变成相等
3. 对于其他数进行一次遍历，分类讨论如果为偶数则选择i与n,奇数则选择1与i

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <map>
#include <vector>
using namespace std;
#define int long long
typedef pair<int, int> PII;
const int N = 2e5 + 10;
int t, n, m, k;
int a[N];
map<string, int> mp1;
vector<string> p;
signed main()
{
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        if (n == 1)
            cout << 0 << endl;
        else
        {
            if (a[1] == a[n])
            {
                int res = a[1];
                cout << n - 2 << endl;
                for (int i = 2; i <= n - 1; i++)
                {
                    int demo = a[i] + res;
                    if (demo % 2 == 0)
                        cout << i << ' ' << n << endl;
                    else
                        cout << 1 << ' ' << i << endl;
                }
            }
            else
            {
                int res = a[1];
                int res1 = a[n];
                int demo = 0;
                if ((res + res1) % 2 == 0)
                    demo = res1;
                else
                    demo = res;
                cout << n - 1 << endl;
                cout << 1 << ' ' << n << endl;
                for (int i = 2; i <= n - 1; i++)
                {
                    int demo1 = a[i] + demo;
                    if (demo1 % 2 == 0)
                        cout << i << ' ' << n << endl;
                    else
                        cout << 1 << ' ' << i << endl;
                }
            }
        }
    }
    return 0;
}