Problem: 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).

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public class Asd {
static PrintWriter w=new PrintWriter(System.out);
static FastScanner s=new FastScanner();
	public static void main(String[] args) {
		
		int t=1;//s.nextInt();
		while(t-->0)
                {
		    solve();
		}
	       w.close();}
        public static void solve()
        {
           String str=s.next();int len=str.length();int prev=0;int flag=4;
           int count=0;
           for(int i=0;i<=str.length()-4;i++)
           {
               String st=str.substring(i,i+4);
               if(st.equals("bear"))
               {
                   if(i==0)
                       count+=(len-i-4+1);
                   else if(i+4==len)
                   {
                       if(prev==0&&flag==4)
                       count+=(i-prev+1);
                       else
                       count+=(i-prev);                        
                   }
                   else
                   {
                       if(prev==0&&flag==4)
                           count+=(i-prev+1)*(len-i-3);
                        else
                         count+=(i-prev)*(len-i-3);
                   }
                   prev=i;flag=5;
               }
           }
           w.println(count);
            
           
                
            
        }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		public String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int[] readArray(int n) {
			int[] a=new int[n]; 
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		int nextInt () {
			return Integer.parseInt(next());
		}
                long nextLong () {
			return Long.parseLong(next());
		}
                double nextDouble () {
			return Double.parseDouble(next());
		}
	}

}