The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.

String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.

Help the bear cope with the given problem.

Input

The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.

Output

Print a single number — the answer to the problem.

Examples

Input

bearbtear


Output

6


Input

bearaabearc


Output

20

Note

In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).

In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
import java.io.*;
import java.util.*;
import java.math.*;

public class Main{
    public static void main(String args[]) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String bear = "bear";
        String s = br.readLine();
        ArrayList<Integer> l = new ArrayList<Integer>();
        ArrayList<Integer> r = new ArrayList<Integer>();
        long ans = 0;
        for(int i = 0 ; i < s.length() ; i++){
            int count = 0;
            for(int j = 0 ; j < bear.length() ; j++){
                if(i+count < s.length() && s.charAt(i+count) == bear.charAt(j))
                    count++;
            }
            if(count == 4) {
                l.add(i); r.add(i + 3);
            }
        }
        for(int i = 0 ; i < l.size() ; i++){
            int left = l.get(i); int right = r.get(i);
            if(i ==0)
                ans += (s.length() - right) * (left+1);
            else
                ans += (s.length() - right) *(left - l.get(i-1));
            
        }
        
        System.out.println(ans);
    }
}