Problem: 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.

Examples

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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main{
    final boolean isFileIO = false; 
    BufferedReader in;
    PrintWriter out;
    StringTokenizer st = new StringTokenizer("");
    String delim = " ";

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.initIO();
        m.solve();
        m.in.close();
        m.out.close();
    }

    public void initIO() throws IOException {
        if(!isFileIO) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt"); 
        }
    }
  
    String nextToken() throws IOException {
        if(!st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken(delim);
    }
 
    String readLine() throws IOException {
        return in.readLine();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    void solve() throws IOException {
        String s = readLine();
        int n = s.length();
        int[][] cnt = new int[n + 1][4];
        for(int i = 1; i <= n; i++) {
            for(int j = 0; j < 4; j++)
                cnt[i][j] = cnt[i - 1][j];
            if(i % 2 == 0) {
                if(s.charAt(i - 1) == 'a')
                    cnt[i][0]++;
                else
                    cnt[i][1]++;
            } else {
                if(s.charAt(i - 1) == 'a')
                    cnt[i][2]++;    
                else
                    cnt[i][3]++;
            }
        }
        long anse = 0;
        long anso = 0;
        for(int i = 1; i <= n; i++) {
            if(s.charAt(i - 1) == 'a') {
                if(i % 2 == 0) {
                    anso += cnt[i][0];
                    anse += cnt[i][2];
                }
                else {
                    anso += cnt[i][2];
                    anse += cnt[i][0];
                }
            } else {
                if(i % 2 == 0) {
                    anso += cnt[i][1];
                    anse += cnt[i][3];
                } else {
                    anso += cnt[i][3];
                    anse += cnt[i][1];
                }
            }
        }
        out.println(String.format("%d %d", anse, anso));
    }


}