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.
'''
# Submitted By M7moud Ala3rj
Don't Copy This Code, CopyRight . [email protected] © 2022-2023 :)
'''
# Problem Name = "Bear and Strings"
# Class: B

import sys
#sys.setrecursionlimit(2147483647)
input = sys.stdin.readline
def print(*args, end='\n', sep=' ') -> None:
    sys.stdout.write(sep.join(map(str, args)) + end)

def Solve():
    s = input().strip()
    xs = [] ; ans=0
    for i in range(len(s)):
        if s[i:i+4]=="bear": xs.append(i)
    i = 0
    for x in xs:
        while i<len(s):
            if x>=i:
                ans+=len(s)-(x+3)
                i+=1
            else: break
    
    print(ans)


if __name__ == "__main__":
    Solve()