CF Round 821 (Div. 2)--C. Parity Shuffle Sorting-Toy模板网

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You are given an array a with n non-negative integers. You can apply the following operation on it.

Choose two indices l and r (1≤l<r≤n).
If al+ar is odd, do ar:=al. If al+ar is even, do al:=ar.

Find any sequence of at most n operations that makes a non-decreasing. It can be proven that it is always possible. Note that you do not have to minimize the number of operations.

An array a1,a2,…,an is non-decreasing if and only if a1≤a2≤…≤an.

Input

The first line contains one integer t (1≤t≤10^5) — the number of test cases.

Each test case consists of two lines. The first line of each test case contains one integer n (1≤n≤10^5) — the length of the array.

The second line of each test case contains n integers a1,a2,…,an (0≤ai≤10^9) — the array itself.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.

Output

For each test case, print one integer m (0≤m≤n), the number of operations, in the first line.

Then print m lines. Each line must contain two integers li,ri, which are the indices you chose in the i-th operation (1≤li<ri≤n).

If there are multiple solutions, print any of them.

input

3
2
7 8
5
1 1000000000 3 0 5
1
0

output

Note

In the second test case, a changes like this:

Select indices 3 and 4. a3+a4=3 is odd, so do a4:=a3. a=[1,1000000000,3,3,5] now.
Select indices 1 and 2. a1+a2=1000000001 is odd, so do a2:=a1. a=[1,1,3,3,5] now, and it is non-decreasing.

In the first and third test cases, a is already non-decreasing.

题意：给定长度为n的数组，在一次操作中，你可以选择 l,r 下标，如果al+ar是奇数，那么ar变成al，否则al变成ar，输出最多操作n次使得原数组单调不减的方案。

解析：我们发现可以从两头开始思考，第一步我们直接选1和n，那么操作之后a1和an就相等了，此时我们需要判断是a1变成an还是an变成a1，然后根据a1和an分两类，奇奇和偶偶，如果是奇奇，那么我们遍历2~n-1，如果它是偶数，那么可以直接选a1跟他匹配，否则就选an跟它匹配，然后就可以使得每个数都相等，操作n-1次，如果是偶偶，匹配反一下就行。

注意：因为规定L!=R，因此当n=1时，我们需要特判一下。文章来源地址https://www.toymoban.com/news/detail-512477.html

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
typedef pair<int,int> PII;
int a[N];
void solve()
{
    int n,op;
    scanf("%d",&n);
    vector<PII> ans;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    if(n!=1) ans.push_back({1,n});//因为l,r需要不同，特判
    if((a[1]+a[n])%2)
    {
        if(a[1]%2) op=1;
        else op=2;
    }else
    {
        if(a[n]%2) op=1;
        else op=2;
    }
    if(op==1)//奇奇操作
    {
        for(int i=2;i<n;i++)
        {
            if(a[i]%2==0) ans.push_back({1,i});
            else ans.push_back({i,n});
        }
    }
    else//偶偶操作
    {
        for(int i=2;i<n;i++)
        {
            if(a[i]%2) ans.push_back({1,i});
            else ans.push_back({i,n});
        }
    }
    printf("%d\n",ans.size());
    for(int i=0;i<ans.size();i++) printf("%d %d\n",ans[i].first,ans[i].second);
}
int main()
{
    int t=1;
    scanf("%d",&t);
    while(t--) solve();
    return 0;
}

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