B. Interesting Subarray

B. Interesting Subarray

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

For an array aa of integers let's denote its maximal element as max(a)max(a), and minimal as min(a)min(a). We will call an array aa of kk integers interesting if max(a)−min(a)≥kmax(a)−min(a)≥k. For example, array [1,3,4,3][1,3,4,3] isn't interesting as max(a)−min(a)=4−1=3<4max(a)−min(a)=4−1=3<4 while array [7,3,0,4,3][7,3,0,4,3] is as max(a)−min(a)=7−0=7≥5max(a)−min(a)=7−0=7≥5.

You are given an array aa of nn integers. Find some interesting nonempty subarray of aa, or tell that it doesn't exist.

An array bb is a subarray of an array aa if bb can be obtained from aa by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. In particular, an array is a subarray of itself.

Input

The first line contains integer number tt (1≤t≤100001≤t≤10000). Then tt test cases follow.

The first line of each test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the length of the array.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109) — the elements of the array.

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, output "NO" in a separate line if there is no interesting nonempty subarray in aa.

Otherwise, output "YES" in a separate line. In the next line, output two integers ll and rr (1≤l≤r≤n1≤l≤r≤n) — bounds of the chosen subarray. If there are multiple answers, print any.

You can print each letter in any case (upper or lower).

Example

input

Copy

3
5
1 2 3 4 5
4
2 0 1 9
2
2019 2020

output

Copy

NO
YES
1 4
NO

Note

In the second test case of the example, one of the interesting subarrays is a=[2,0,1,9]a=[2,0,1,9]: max(a)−min(a)=9−0=9≥4max(a)−min(a)=9−0=9≥4.

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钻牛角尖的题，k越小越好，那就取2，判断连续两个数字是否是绝对值相差2

# include
# include
# include
# include
# include
using namespace std;
typedef long long int ll;
 
int main ()
{
 
    int t;
 
    cin>>t;
 
    while(t--)
    {
        int n,pre,x,flag=0,ans1,ans2;
        cin>>n>>pre;
 
        for(int i=2;i<=n;i++)
        {
            cin>>x;
 
            if(abs(x-pre)>=2)
            {
                ans1=i-1;
                ans2=i;
                flag=1;
            }
            pre=x;
        }
 
        if(flag)
        {
            cout<<"YES"<
            cout<" "<
        }
        else
        {
            cout<<"NO"<
        }
    }
 
    return 0;
}