根据题意可知只要存在前后差大于等于2的两个元素即可

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<cstdlib>
//#include<windows.h>
#define fi first
#define se second
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define mes(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int N = 1e6+10;
const int INF = 1e9+5;
const int inf = - INF;
const int mod = 1e9+7;
const double pi = acos(-1.0);
int a[N];
int main(){
    std::ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--){
        int n;
        int flag=0;
        cin>>n;
        rep(i,1,n)
            cin>>a[i];
        rep(i,1,n-1)
            if(abs(a[i+1]-a[i])>=2){
                cout<<"YES"<<endl;
                cout<<i<<' '<<i+1<<endl;
                flag=1;
                break;
            }
        if(!flag) cout<<"NO"<<endl;
    }
    //system("pause");
    return 0;
}