You are given a string s, consisting of n letters, each letter is either 'a' or 'b'. The letters in the string are numbered from 1 to n.

s[l; r] is a continuous substring of letters from index l to r of the string inclusive. 

A string is called balanced if the number of letters 'a' in it is equal to the number of letters 'b'. For example, strings "baba" and "aabbab" are balanced and strings "aaab" and "b" are not.

Find any non-empty balanced substring s[l; r] of string s. Print its l and r (1 ≤ l ≤ r ≤ n). If there is no such substring, then print -1 -1.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Then the descriptions of t testcases follow.

The first line of the testcase contains a single integer n (1 ≤ n ≤ 50) — the length of the string.

The second line of the testcase contains a string s, consisting of n letters, each letter is either 'a' or 'b'.

Output

For each testcase print two integers. If there exists a non-empty balanced substring s[l; r], then print l r (1 ≤ l ≤ r ≤ n). Otherwise, print -1 -1.

Example

Input


4
1
a
6
abbaba
6
abbaba
9
babbabbaa


Output


-1 -1
1 6
3 6
2 5

Note

In the first testcase there are no non-empty balanced subtrings.

In the second and third testcases there are multiple balanced substrings, including the entire string "abbaba" and substring "baba".

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;


public class q_1 {
    static Scanner sc;
    static FastReader fr;
    public static void main(String[] args){
        sc = new Scanner(System.in);
        fr = new FastReader();
        int n = fr.nextInt();

        for(int test=0; test<n; test++){
            int len = fr.nextInt();
            String s = fr.nextLine();
            char[] chars = s.toCharArray();
            boolean ans = false;
            for(int j=1; j<len; j++){
                if(chars[j-1]!=chars[j]){
                    System.out.println(j+" "+(j+1));
                    ans = true;
                    break;
                }
            }
            if(!ans){
                System.out.println(-1+" "+(-1));
            }
        }
    }

    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static Integer[] getIntArrayFromSingleLine(){
        List<Integer> list = new LinkedList<>();
        String[] str = fr.nextLine().split(" ");
        for(String s: str){
            list.add(Integer.parseInt(s));
        }
        Integer[] arr = new Integer[list.size()];
        return list.toArray(arr);
    }

    public static Long[] getLongArrayFromSingleLine(){
        List<Long> list = new LinkedList<>();
        String[] str = fr.nextLine().split(" ");
        for(String s: str){
            list.add(Long.parseLong(s));
        }
        Long[] longs = new Long[list.size()];
        return list.toArray(longs);
    }

    public static Double[] getDoubleArrayFromSingleLine(){
        List<Double> list = new LinkedList<>();
        String[] str= fr.nextLine().split(" ");
        for(String s: str){
            list.add(Double.parseDouble(s));
        }
        Double[] doubles = new Double[list.size()];
        return list.toArray(doubles);
    }

    public static String[] getStringArrayFromSingleLine(){
        return fr.nextLine().split(" ");
    }

    public static int[] getArray(int n){
        int[] arr = new int[n];
        for(int i=0; i<n; i++){
            arr[i] = fr.nextInt();
        }
        return arr;
    }

    public static long[] getLongArray(int n){
        long[] res = new long[n];
        for(int i=0; i<n; i++){
            res[i] = fr.nextLong();
        }
        return res;
    }

    public static double[] getDoubleArray(int n){
        double[] arr = new double[n];
        for(int i=0; i<n ; i++){
            arr[i] = fr.nextDouble();
        }
        return arr;
    }

    public static char[] getCharArray(){
        return fr.next().toCharArray();
    }

    public static String[] getStringArray(int n){
        String[] strings = new String[n];
        for(int i=0; i<n; i++){
            strings[i] = fr.next();
        }
        return strings;
    }

    // Get GCD
    public static int getGCD(int a, int b){
        return (a==0 ? b : getGCD(b%a, a));
    }

    public static int getGCD(int[] arr){
        int n = arr.length;
        if(n==0){
            return -1;
        }
        int res = arr[0];
        for(int i=1; i<n; i++){
            res = getGCD(res, arr[i]);
        }
        return res;
    }

    // Get all prime numbers till N using seive of Eratosthenes
    public static List<Integer> getPrimesTill(int n){
        boolean[] arr = new boolean[n+1];
        List<Integer> primes = new LinkedList<>();
        for(int i=2; i<=n; i++){
            if(!arr[i]){
                primes.add(i);
                for(long j=(i*i); j<=n; j+=i){
                    arr[(int)j] = true;
                }
            }
        }
        return primes;
    }


    // This method returns list of PrimeFactors of a number in O(sqrt(N));
    public static Map<Integer,Integer> getPrimeFactors(long n){
        Map<Integer,Integer> res = new HashMap<>();
        int sqrt = (int) Math.sqrt(n);
        if(n%2 == 0){
            int c = 0;
            while(n%2==0){
                c++;
                n/=2;
            }
            res.put(2,c);
        }
        for(int i=3; i<=sqrt; i+=2){
            if(n%i == 0) {
                int c = 0;
                while (n % i == 0) {
                    c++;
                    n /= i;
                }
                res.put(i,c);
            }
        }
        if(n!=1){
            res.put((int)n,1);
        }
        return res;
    }

}