human
Can you solve the below in PYTHON?
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".

=====
Give me a Python program to find any non-empty balanced substring s[l?
assistant
#Here is Python program to find any non-empty balanced substring s[l.

import sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
interactive = False
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

class InputData:
    def __init__(self):
        self.lines = readAll().split('\n')
        self.n = len(self.lines)
        self.ii = -1
    def input(self):
        self.ii += 1
        assert self.ii < self.n
        return self.lines[self.ii]

flush = sys.stdout.flush
if interactive and not testing:
    input = sys.stdin.readline
else:
    inputData = InputData()
    input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    return(list(input()))
def strin():
    return input()
def lout(l, sep="\n", toStr=True):
    print(sep.join(map(str, l) if toStr else l))
def dout(*args, **kargs):
    if not testing: return
    if args: print(args[0] if len(args)==1 else args)
    if kargs: print([(k,v) for k,v in kargs.items()])
def ask(q):
    sys.stdout.write(str(q)+'\n')
    flush()
    return intin()
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
# from collections import defaultdict as ddict, Counter
# from heapq import *
# from Queue import Queue

def main():
    n = intin()
    a = strin()
    for i in xrange(n-1):
        for j in xrange(i+1,n):
            cnt = 0
            for k in xrange(i,j+1):
                if a[k] == 'a':
                    cnt += 1
                else:
                    cnt -= 1
            if cnt == 0:
                return '%d %d' % (i+1,j+1)
    return '-1 -1'

anss = []
for _ in xrange(intin()):
    anss.append(main())
    # anss.append("YES" if main() else "NO")
lout(anss)

if testing:
    sys.stdout = cmd
    print(int(round(time() * 1000))  - start_time)