/*
    written by Pankaj Kumar.
    country:-INDIA
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll ;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define line cout<<endl;
#define pb push_back
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} line;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())
// datatype definination
#define ordered_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>


class Codeforces{
    private:
        // read only variable
        const ll INF=1e18;
        const ll mod1=1e9+7;
        const ll mod2=998244353;


    public:
        Codeforces(){

        }

        ll power(ll x,ll y){
            ll result=1;
            while(y>0){
                if(y&1){
                    result*=x;
                }
                y>>=1;
                x*=x;
            }
            return result;
        }

        ll power(ll x,ll y,ll mod){
            ll result=1;
            x%=mod;
            while(y>0){
                if(y&1){
                    result*=x;
                    result%=mod;
                }
                y>>=1;
                x*=x;
                x%=mod;
            }
            return result;
        }

        ll str_to_num(string s){
            stringstream pk(s);
            ll num;
            pk>>num;
            return num;
        }

        string num_to_str(ll num){
            return to_string(num);
        }
        // Techniques :
        // divide into cases, brute force, pattern finding
        // sort, greedy, binary search, two pointer
        // transform into graph

        // Experience :
        // Cp is nothing but only observation and mathematics.
        ll solve(){
            ll n;
            string s;
            cin>>n>>s;
            if(n&1){
                no
                return 0;
            }
            else{
                ll one=0,zero=0;
                for(auto x:s){
                    if(x=='0'){
                        zero++;
                    }
                    else{
                        one++;
                    }
                }
                ll required=n/2;
                if(zero==required){
                    yes
                    cout<<1<<" "<<n<<endl;
                    return 0;
                }
                if(one>zero){
                    ll diff=one-zero;
                    diff/=2;
                    ll count=0;
                    // here we need a range l,r such that it have x zero and x+diff one
                    for(ll i=0;i<n;i++){
                        if(s[i]=='1'){
                            count++;
                        }
                        else{
                            count--;
                        }
                        if(count==diff){
                            yes
                            cout<<1<<" "<<i+1<<endl;
                            return 0;
                        }
                    }
                    count=0;
                    for(ll i=n-1;i>=0;i--){
                        if(s[i]=='1'){
                            count++;
                        }
                        else{
                            count--;
                        }
                        if(count==diff){
                            yes
                            cout<<i+1<<" "<<n<<endl;
                            return 0;
                        }
                    }
                }
                else{
                    ll diff=zero-one;
                    diff/=2;
                    // here we need a range l,r such that it have x one and x+diff zero
                    ll count=0;
                    for(ll i=0;i<n;i++){
                        if(s[i]=='0'){
                            count++;
                        }
                        else{
                            count--;
                        }
                        if(count==diff){
                            yes
                            cout<<1<<" "<<i+1<<endl;
                            return 0;
                        }
                    }
                    count=0;
                    for(ll i=n-1;i>=0;i--){
                        if(s[i]=='0'){
                            count++;
                        }
                        else{
                            count--;
                        }
                        if(count==diff){
                            yes
                            cout<<i+1<<" "<<n<<endl;
                            return 0;
                        }
                    }
                }

            }
            no
            return 0;
        }
};


/*  --------------------MAIN PROGRAM----------------------------*/

int main()
{
    speed;
    /* #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif */
    ll TestCase=1;
    cin>>TestCase;
    while(TestCase--){
        Codeforces cf;
        cf.solve();
    }
}
/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/