Codeforces 451 D. Count Good Substrings

D. Count Good Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. For example, "aabba" is good, because after the merging step it will become "aba".

Given a string, you have to find two values:

1. the number of good substrings of even length;
2. the number of good substrings of odd length.

Input

The first line of the input contains a single string of length n (1 ≤ n ≤ 105). Each character of the string will be either 'a' or 'b'.

Output

Print two space-separated integers: the number of good substrings of even length and the number of good substrings of odd length.

Sample test(s)

input

bb

output

1 2

input

baab

output

2 4

input

babb

output

2 5

input

babaa

output

2 7

Note

In example 1, there are three good substrings ("b", "b", and "bb"). One of them has even length and two of them have odd length.

In example 2, there are six good substrings (i.e. "b", "a", "a", "b", "aa", "baab"). Two of them have even length and four of them have odd length.

In example 3, there are seven good substrings (i.e. "b", "a", "b", "b", "bb", "bab", "babb"). Two of them have even length and five of them have odd length.

Definitions

A substring s[l, r] (1 ≤ l ≤ r ≤ n) of string s = s1s2... sn is string slsl + 1... sr.

A string s = s1s2... sn is a palindrome if it is equal to string snsn - 1... s1.

import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
		String str=in.nextLine();
		long a0=0,a1=0,b0=0,b1=0,d=0;
		for(int i=0;i<str.length();i++)
		{
			if(str.charAt(i)=='a')
			{
				if(i%2==0)
				{
					a0++;
					d+=a0;
				}
				else
				{
					a1++;
					d+=a1;
				}
			}
			else if(str.charAt(i)=='b')
			{
				if(i%2==0)
				{
					b0++;
					d+=b0;
				}
				else
				{
					b1++;
					d+=b1;
				}
			}
		}
		System.out.println((a0*a1+b0*b1)+" "+d);
	}
}