The bear has a string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (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*) = *s*<sub class="lower-index">*i*</sub>*s*<sub class="lower-index">*i* + 1</sub>... *s*<sub class="lower-index">*j*</sub> 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 *s*<sub class="lower-index">*k*</sub> = *b*, *s*<sub class="lower-index">*k* + 1</sub> = *e*, *s*<sub class="lower-index">*k* + 2</sub> = *a*, *s*<sub class="lower-index">*k* + 3</sub> = *r*.

Help the bear cope with the given problem.
The first line contains a non-empty string *s* (1 ≤ |*s*| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
import logging
import copy
import sys
import re

         
logging.basicConfig(stream=sys.stderr, level=logging.DEBUG)

# def subSolve(text):
#     s = 0
#     for m in re.finditer('bear', text):
#         s += (len(text) - m.end() + 1)
#     return s
    
def solve(text):
    s = 0

    startIdx = []
    
    for i in range(len(text)- 4):
        if text[i:i+4] == "bear":
            startIdx.append(i)

    for idx, i in enumerate(startIdx):
        if idx == 0:
            s += ((i+1) * (len(text) - (i+4) + 1))
        else :
            s += ( (i - startIdx[idx-1] )  * (len(text) - (i+4) + 1))
        
    
    return s

def main():
    firstLine = input()
    print(solve(firstLine.strip()))

def log(*message):
    logging.debug(message)
    
if __name__ == "__main__":
    main()
