Problem: Given a string t, we will call it unbalanced if and only if the length of t is at least 2, and more than half of the letters in t are the same. For example, both `voodoo` and `melee` are unbalanced, while neither `noon` nor `a` is.

You are given a string s consisting of lowercase letters. Determine if there exists a (contiguous) substring of s that is unbalanced. If the answer is positive, show a position where such a substring occurs in s.

Constraints

* 2 ≦ |s| ≦ 10^5
* s consists of lowercase letters.

Input

The input is given from Standard Input in the following format:


s


Output

If there exists no unbalanced substring of s, print `-1 -1`.

If there exists an unbalanced substring of s, let one such substring be s_a s_{a+1} ... s_{b} (1 ≦ a < b ≦ |s|), and print `a b`. If there exists more than one such substring, any of them will be accepted.

Examples

Input

needed


Output

2 5


Input

atcoder


Output

-1 -1

Solution (in PYTHON):

Solution:
import bisect
import sys
import math
import itertools
sys.setrecursionlimit(10000)

INF = float('inf')

# input macro
def i():
    return int(raw_input())
def ii():
    return map(int,raw_input().split(" "))
def s():
    return raw_input()
def ss():
    return raw_input().split(" ")
def slist():
    return list(raw_input())
#

def join(s):
    return ''.join(s)

#iterate macro
def piter(n,m):
    return itertools.permutations(n,m)
def citer(n,m):
    return itertools.combinations(n,m)

#modulo macro
def modc(a,b,m):
    c = 1
    for i in xrange(b):
        c = c * (a - i) % m
        c = c * modinv(i + 1,m) % m
    return c
 
def gcd(a, b):
    (x, lastx) = (0, 1)
    (y, lasty) = (1, 0)
    while b != 0:
        q = a // b
        (a, b) = (b, a % b)
        (x, lastx) = (lastx - q * x, x)
        (y, lasty) = (lasty - q * y, y)
    return (lastx, lasty, a)
 
def modinv(a, m):
    (inv, q, gcd_val) = gcd(a, m)
    return inv % m

#bisect macro
def index(a, x):
    #Locate the leftmost value exactly equal to x
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    return -1

#memoize macro
def memoize(f):
    cache = {}
    def helper(*args):
        if args not in cache:
            cache[(args)] = f(*args)
        return cache[args]
    return helper

@memoize
def nck(a,b,m):
    b = min([a-b,b])
    if (b>a or b<0 or a<0):
        return 0
    elif a == 0:
        return 1
    return (nck(a-1,b-1,m)+nck(a-1,b,m)) % m

###########

s=slist()
n=len(s)
if s[n-2]==s[n-1]:
    print n-1,n
else:
    for i in range(n-2):
    
        if s[i]==s[i+1]:
            print i+1,i+2
            break
        elif s[i]==s[i+2]:
            print i+1,i+3
            break
    else:
        print -1,-1